In this answer there're some examples of well-defined and undefined expressions. I'm particularly interested in two of them:
(6) i = i++ + 1; // Undefined Behaviour
(7) i = ++i + 1; // Well-defined Behaviour
This means that there's a difference between pre-increment and post-increment in terms of sequence points and well defined /unspecified/undefined behavior, but I don't understand where this difference comes from.
In standard draft (N4618) there's an example of code ([intro.execution], pt 18)
i = i++ + 1; // the value of i is incremented
i = i++ + i; // the behavior is undefined
Which, as far as I understand, means that expression i = i++ + 1 should be well-defined and the value of a variable i should increase by 1 as the result of this expression. However, this code run in MSVS 2015 increases i by 2.
So, what happens in the expression i = i++ + 1? Is it well-defined, undefined, implementation-defined or unspecified behavior? And is there any difference between pre-increment and post-increment in this and similar expressions in terms of sequence points and UB, as stated in the original answer? And why Visual Studio shows the behavior which is different from written in standard?
Please also note that I'm primarily interested in modern c++ (14/17).
What happens in the expression i = i++ + 1? Is it well-defined, undefined, implementation defined or unspecified behaviour?
This exact example is given in the standard, how lucky are we?
N4296 1.9.15 [intro.execution]
i = i++ + 1; // the behavior is undefined
Of course, we'd like to know why too. The following standard quote appears to be relevant here:
N4296 1.9.15 [intro.execution]
[ ... ] The value computations of the operands of an operator are sequenced
before the value computation of the result of the operator. [ ... ]
This tells us that the sum will occur before the assignment (duh, how else does it know what to assign!), but it doesn't guarantee that the increment will occur before or after the assignment, now we're in murky water...
N4296 1.9.15 [intro.execution]
[ ... ] If a side effect on a scalar object is unsequenced relative to either
another side effect on the same scalar object or a value computation
using the value of the same scalar object, and they are not
potentially concurrent (1.10), the behavior is undefined. [ ... ]
The assignment operator has a side effect on the value of i, which means we have two side effects (the other is the assignment performed by i++) on the same scalar object, which are unsequenced, which is undefined.
Why does Visual Studio show the behavior which is different from written in standard?
It doesn't. The standard says it's undefined, which means it can do anything from what you wanted to something completely different, it just so happens that this is the behaviour that got spat out by the compiler!
i = ++i + 1; // Well-defined Behaviour
This means that there's a difference between preincrement and postincrement in terms of sequence points and well defined / unspecified / undefined behavior, but I don't understand where this difference comes from.
I believe that post is incorrect in several ways. Quoting the same section as they do, emphasis mine:
C++11 1.9/15
The value computations of the operands of an
operator are sequenced before the value computation of the result of the operator. If a side effect on a scalar
object is unsequenced relative to either another side effect on the same scalar object or a value computation
using the value of the same scalar object, the behavior is undefined.
Then the assignment operator:
C++11 5.17
In all cases, the assignment is sequenced after the value computation of the right and left operands, and before the value computation of the assignment expression.
Notable, the value computation or the right and left operands is not sequenced. (This has been explicitly spelled out in C11 1), which otherwise has a completely identical text as C++11.)
Meaning that in the expression i = ++i + 1; the side effect of ++i is unsequenced in relation to the value computation of the left operand i. Thus it is undefined behavior as per 1.9/15. And the UB has nothing to do with the assignment side-effect at all.
As for the expression i = i++ + 1;, the side-effect of assignment is, as per C++11, explicitly sequenced after the value computations, but before the value computation of the expression as whole. The value computation of i++ is not an issue as per 5.2.6 "The value computation of the ++ expression is sequenced before the modification of the operand object". As per the nature of the postfix ++, the side effect of updating i++ must be sequenced after the value computation of the whole expression. It is well-defined behavior, as far as I can tell.
Correct text should therefore be
(6) i = i++ + 1; // Well-defined Behaviour
(7) i = ++i + 1; // Undefined Behaviour
Apparently there was an incorrect example in C++11 1.9/15 i = i++ + 1; // the behavior is undefined which has been corrected in later versions of the standard.
NOTE: None of this has the slightest to do with the change of wording about sequence points!
1) C11 6.5.16/3
The side effect of updating the stored value of the
left operand is sequenced after the value computations of the left and
right operands. The evaluations of the operands are unsequenced.
I have been fooling around with some code and saw something that I don't understand the "why" of.
int i = 6;
int j;
int *ptr = &i;
int *ptr1 = &j
j = i++;
//now j == 6 and i == 7. Straightforward.
What if you put the operator on the left side of the equals sign?
++ptr = ptr1;
is equivalent to
(ptr = ptr + 1) = ptr1;
whereas
ptr++ = ptr1;
is equivalent to
ptr = ptr + 1 = ptr1;
The postfix runs a compilation error and I get it. You've got a constant "ptr + 1" on the left side of an assignment operator. Fair enough.
The prefix one compiles and WORKS in C++. Yes, I understand it's messy and you're dealing with unallocated memory, but it works and compiles. In C this does not compile, returning the same error as the postfix "lvalue required as left operand of assignment". This happens no matter how it's written, expanded out with two "=" operators or with the "++ptr" syntax.
What is the difference between how C handles such an assignment and how C++ handles it?
In both C and C++, the result of x++ is an rvalue, so you can't assign to it.
In C, ++x is equivalent to x += 1 (C standard §6.5.3.1/p2; all C standard cites are to WG14 N1570). In C++, ++x is equivalent to x += 1 if x is not a bool (C++ standard §5.3.2 [expr.pre.incr]/p1; all C++ standard cites are to WG21 N3936).
In C, the result of an assignment expression is an rvalue (C standard §6.5.16/p3):
An assignment operator stores a value in the object designated by the
left operand. An assignment expression has the value of the left
operand after the assignment, but is not an lvalue.
Because it's not an lvalue, you can't assign to it: (C standard §6.5.16/p2 - note that this is a constraint)
An assignment operator shall have a modifiable lvalue as its left
operand.
In C++, the result of an assignment expression is an lvalue (C++ standard §5.17 [expr.ass]/p1):
The assignment operator (=) and the compound assignment operators all
group right-to-left. All require a modifiable lvalue as their left
operand and return an lvalue referring to the left operand.
So ++ptr = ptr1; is a diagnosable constraint violation in C, but does not violate any diagnosable rule in C++.
However, pre-C++11, ++ptr = ptr1; has undefined behavior, as it modifies ptr twice between two adjacent sequence points.
In C++11, the behavior of ++ptr = ptr1 becomes well defined. It's clearer if we rewrite it as
(ptr += 1) = ptr1;
Since C++11, the C++ standard provides that (§5.17 [expr.ass]/p1)
In all cases, the assignment is sequenced after the value computation
of the right and left operands, and before the value computation of
the assignment expression. With respect to an
indeterminately-sequenced function call, the operation of a compound
assignment is a single evaluation.
So the assignment performed by the = is sequenced after the value computation of ptr += 1 and ptr1. The assignment performed by the += is sequenced before the value computation of ptr += 1, and all value computations required by the += are necessarily sequenced before that assignment. Thus, the sequencing here is well-defined and there is no undefined behavior.
In C the result of pre and post increment are rvalues and we can not assign to an rvalue, we need an lvalue(also see: Understanding lvalues and rvalues in C and C++) . We can see by going to the draft C11 standard section 6.5.2.4 Postfix increment and decrement operators which says (emphasis mine going forward):
The result of the postfix ++ operator is the value of the
operand. [...] See the discussions of additive operators and compound
assignment for information on constraints, types, and conversions and
the effects of operations on pointers. [...]
So the result of post-increment is a value which is synonymous for rvalue and we can confirm this by going to section 6.5.16 Assignment operators which the paragraph above points us to for further understanding of constraints and results, it says:
[...] An assignment expression has the value of the left operand after the
assignment, but is not an lvalue.[...]
which further confirms the result of post-increment is not an lvalue.
For pre-increment we can see from section 6.5.3.1 Prefix increment and decrement operators which says:
[...]See the discussions of additive operators and compound assignment for
information on constraints, types, side effects, and conversions and
the effects of operations on pointers.
also points back to 6.5.16 like post-increment does and therefore the result of pre-increment in C is also not an lvalue.
In C++ post-increment is also an rvalue, more specifically a prvalue we can confirm this by going to section 5.2.6 Increment and decrement which says:
[...]The result is a prvalue. The type of the result is the cv-unqualified
version of the type of the operand[...]
With respect to pre-increment C and C++ differ. In C the result is an rvalue while in C++ the result is a lvalue which explains why ++ptr = ptr1; works in C++ but not C.
For C++ this is covered in section 5.3.2 Increment and decrement which says:
[...]The result is the updated operand; it is an lvalue, and it is a
bit-field if the operand is a bit-field.[...]
To understand whether:
++ptr = ptr1;
is well defined or not in C++ we need two different approaches one for pre C++11 and one for C++11.
Pre C++11 this expression invokes undefined behavior, since it is modifying the object more than once within the same sequence point. We can see this by going to a Pre C++11 draft standard section 5 Expressions which says:
Except where noted, the order of evaluation of operands of individual
operators and subexpressions of individual expressions, and the order
in which side effects take place, is unspecified.57) Between the
previous and next sequence point a scalar object shall have its stored
value modified at most once by the evaluation of an expression.
Furthermore, the prior value shall be accessed only to determine the
value to be stored. The requirements of this paragraph shall be met
for each allowable ordering of the subexpressions of a full
expression; otherwise the behavior is undefined. [ Example:
i = v[i ++]; / / the behavior is undefined
i = 7 , i++ , i ++; / / i becomes 9
i = ++ i + 1; / / the behavior is undefined
i = i + 1; / / the value of i is incremented
—end example ]
We are incrementing ptr and then subsequently assigning to it, which is two modifications and in this case the sequence point occurs at the end of the expression after the ;.
For C+11, we should go to defect report 637: Sequencing rules and example disagree which was the defect report that resulted in:
i = ++i + 1;
becoming well defined behavior in C++11 whereas prior to C++11 this was undefined behavior. The explanation in this report is one of best I have even seen and reading it many times was enlightening and helped me understand many concepts in a new light.
The logic that lead to this expression becoming well defined behavior goes as follows:
The assignment side-effect is required to be sequenced after the value computations of both its LHS and RHS (5.17 [expr.ass] paragraph 1).
The LHS (i) is an lvalue, so its value computation involves computing the address of i.
In order to value-compute the RHS (++i + 1), it is necessary to first value-compute the lvalue expression ++i and then do an lvalue-to-rvalue conversion on the result. This guarantees that the incrementation side-effect is sequenced before the computation of the addition operation, which in turn is sequenced before the assignment side effect. In other words, it yields a well-defined order and final value for this expression.
The logic is somewhat similar for:
++ptr = ptr1;
The value computations of the LHS and RHS are sequenced before the assignment side-effect.
The RHS is an lvalue, so its value computation involves computing the address of ptr1.
In order to value-compute the LHS (++ptr), it is necessary to first value-compute the lvalue expression ++ptr and then do an lvalue-to-rvalue conversion on the result. This guarantees that the incrementation side-effect is sequenced before the assignment side effect. In other words, it yields a well-defined order and final value for this expression.
Note
The OP said:
Yes, I understand it's messy and you're dealing with unallocated
memory, but it works and compiles.
Pointers to non-array objects are considered arrays of size one for additive operators, I am going to quote the draft C++ standard but C11 has almost the exact same text. From section 5.7 Additive operators:
For the purposes of these operators, a pointer to a nonarray object
behaves the same as a pointer to the first element of an array of
length one with the type of the object as its element type.
and further tells us pointing one past the end of an array is valid as long as you don't dereference the pointer:
[...]If both the pointer operand and the result point to elements of
the same array object, or one past the last element of the array
object, the evaluation shall not produce an overflow; otherwise, the
behavior is undefined.
so:
++ptr ;
is still a valid pointer.
I have been fooling around with some code and saw something that I don't understand the "why" of.
int i = 6;
int j;
int *ptr = &i;
int *ptr1 = &j
j = i++;
//now j == 6 and i == 7. Straightforward.
What if you put the operator on the left side of the equals sign?
++ptr = ptr1;
is equivalent to
(ptr = ptr + 1) = ptr1;
whereas
ptr++ = ptr1;
is equivalent to
ptr = ptr + 1 = ptr1;
The postfix runs a compilation error and I get it. You've got a constant "ptr + 1" on the left side of an assignment operator. Fair enough.
The prefix one compiles and WORKS in C++. Yes, I understand it's messy and you're dealing with unallocated memory, but it works and compiles. In C this does not compile, returning the same error as the postfix "lvalue required as left operand of assignment". This happens no matter how it's written, expanded out with two "=" operators or with the "++ptr" syntax.
What is the difference between how C handles such an assignment and how C++ handles it?
In both C and C++, the result of x++ is an rvalue, so you can't assign to it.
In C, ++x is equivalent to x += 1 (C standard §6.5.3.1/p2; all C standard cites are to WG14 N1570). In C++, ++x is equivalent to x += 1 if x is not a bool (C++ standard §5.3.2 [expr.pre.incr]/p1; all C++ standard cites are to WG21 N3936).
In C, the result of an assignment expression is an rvalue (C standard §6.5.16/p3):
An assignment operator stores a value in the object designated by the
left operand. An assignment expression has the value of the left
operand after the assignment, but is not an lvalue.
Because it's not an lvalue, you can't assign to it: (C standard §6.5.16/p2 - note that this is a constraint)
An assignment operator shall have a modifiable lvalue as its left
operand.
In C++, the result of an assignment expression is an lvalue (C++ standard §5.17 [expr.ass]/p1):
The assignment operator (=) and the compound assignment operators all
group right-to-left. All require a modifiable lvalue as their left
operand and return an lvalue referring to the left operand.
So ++ptr = ptr1; is a diagnosable constraint violation in C, but does not violate any diagnosable rule in C++.
However, pre-C++11, ++ptr = ptr1; has undefined behavior, as it modifies ptr twice between two adjacent sequence points.
In C++11, the behavior of ++ptr = ptr1 becomes well defined. It's clearer if we rewrite it as
(ptr += 1) = ptr1;
Since C++11, the C++ standard provides that (§5.17 [expr.ass]/p1)
In all cases, the assignment is sequenced after the value computation
of the right and left operands, and before the value computation of
the assignment expression. With respect to an
indeterminately-sequenced function call, the operation of a compound
assignment is a single evaluation.
So the assignment performed by the = is sequenced after the value computation of ptr += 1 and ptr1. The assignment performed by the += is sequenced before the value computation of ptr += 1, and all value computations required by the += are necessarily sequenced before that assignment. Thus, the sequencing here is well-defined and there is no undefined behavior.
In C the result of pre and post increment are rvalues and we can not assign to an rvalue, we need an lvalue(also see: Understanding lvalues and rvalues in C and C++) . We can see by going to the draft C11 standard section 6.5.2.4 Postfix increment and decrement operators which says (emphasis mine going forward):
The result of the postfix ++ operator is the value of the
operand. [...] See the discussions of additive operators and compound
assignment for information on constraints, types, and conversions and
the effects of operations on pointers. [...]
So the result of post-increment is a value which is synonymous for rvalue and we can confirm this by going to section 6.5.16 Assignment operators which the paragraph above points us to for further understanding of constraints and results, it says:
[...] An assignment expression has the value of the left operand after the
assignment, but is not an lvalue.[...]
which further confirms the result of post-increment is not an lvalue.
For pre-increment we can see from section 6.5.3.1 Prefix increment and decrement operators which says:
[...]See the discussions of additive operators and compound assignment for
information on constraints, types, side effects, and conversions and
the effects of operations on pointers.
also points back to 6.5.16 like post-increment does and therefore the result of pre-increment in C is also not an lvalue.
In C++ post-increment is also an rvalue, more specifically a prvalue we can confirm this by going to section 5.2.6 Increment and decrement which says:
[...]The result is a prvalue. The type of the result is the cv-unqualified
version of the type of the operand[...]
With respect to pre-increment C and C++ differ. In C the result is an rvalue while in C++ the result is a lvalue which explains why ++ptr = ptr1; works in C++ but not C.
For C++ this is covered in section 5.3.2 Increment and decrement which says:
[...]The result is the updated operand; it is an lvalue, and it is a
bit-field if the operand is a bit-field.[...]
To understand whether:
++ptr = ptr1;
is well defined or not in C++ we need two different approaches one for pre C++11 and one for C++11.
Pre C++11 this expression invokes undefined behavior, since it is modifying the object more than once within the same sequence point. We can see this by going to a Pre C++11 draft standard section 5 Expressions which says:
Except where noted, the order of evaluation of operands of individual
operators and subexpressions of individual expressions, and the order
in which side effects take place, is unspecified.57) Between the
previous and next sequence point a scalar object shall have its stored
value modified at most once by the evaluation of an expression.
Furthermore, the prior value shall be accessed only to determine the
value to be stored. The requirements of this paragraph shall be met
for each allowable ordering of the subexpressions of a full
expression; otherwise the behavior is undefined. [ Example:
i = v[i ++]; / / the behavior is undefined
i = 7 , i++ , i ++; / / i becomes 9
i = ++ i + 1; / / the behavior is undefined
i = i + 1; / / the value of i is incremented
—end example ]
We are incrementing ptr and then subsequently assigning to it, which is two modifications and in this case the sequence point occurs at the end of the expression after the ;.
For C+11, we should go to defect report 637: Sequencing rules and example disagree which was the defect report that resulted in:
i = ++i + 1;
becoming well defined behavior in C++11 whereas prior to C++11 this was undefined behavior. The explanation in this report is one of best I have even seen and reading it many times was enlightening and helped me understand many concepts in a new light.
The logic that lead to this expression becoming well defined behavior goes as follows:
The assignment side-effect is required to be sequenced after the value computations of both its LHS and RHS (5.17 [expr.ass] paragraph 1).
The LHS (i) is an lvalue, so its value computation involves computing the address of i.
In order to value-compute the RHS (++i + 1), it is necessary to first value-compute the lvalue expression ++i and then do an lvalue-to-rvalue conversion on the result. This guarantees that the incrementation side-effect is sequenced before the computation of the addition operation, which in turn is sequenced before the assignment side effect. In other words, it yields a well-defined order and final value for this expression.
The logic is somewhat similar for:
++ptr = ptr1;
The value computations of the LHS and RHS are sequenced before the assignment side-effect.
The RHS is an lvalue, so its value computation involves computing the address of ptr1.
In order to value-compute the LHS (++ptr), it is necessary to first value-compute the lvalue expression ++ptr and then do an lvalue-to-rvalue conversion on the result. This guarantees that the incrementation side-effect is sequenced before the assignment side effect. In other words, it yields a well-defined order and final value for this expression.
Note
The OP said:
Yes, I understand it's messy and you're dealing with unallocated
memory, but it works and compiles.
Pointers to non-array objects are considered arrays of size one for additive operators, I am going to quote the draft C++ standard but C11 has almost the exact same text. From section 5.7 Additive operators:
For the purposes of these operators, a pointer to a nonarray object
behaves the same as a pointer to the first element of an array of
length one with the type of the object as its element type.
and further tells us pointing one past the end of an array is valid as long as you don't dereference the pointer:
[...]If both the pointer operand and the result point to elements of
the same array object, or one past the last element of the array
object, the evaluation shall not produce an overflow; otherwise, the
behavior is undefined.
so:
++ptr ;
is still a valid pointer.
This question came up while I was reading (the answers to) So why is i = ++i + 1 well-defined in C++11?
I gather that the subtle explanation is that (1) the expression ++i returns an lvalue but + takes prvalues as operands, so a conversion from lvalue to prvalue must be performed; this involves obtaining the current value of that lvalue (rather than one more than the old value of i) and must therefore be sequenced after the side effect from the increment (i.e., updating i) (2) the LHS of the assignment is also an lvalue, so its value evaluation does not involve fetching the current value of i; while this value computation is unsequenced w.r.t. the value computation of the RHS, this poses no problem (3) the value computation of the assignment itself involves updating i (again), but is sequenced after the value computation of its RHS, and hence after the prvious update to i; no problem.
Fine, so there is no UB there. Now my question is what if one changed the assigment operator from = to += (or a similar operator).
Does the evaluation of the expression i += ++i + 1 lead to undefined behavior?
As I see it, the standard seems to contradict itself here. Since the LHS of += is still an lvalue (and its RHS still a prvalue), the same reasoning as above applies as far as (1) and (2) are concerned; there is no undefined behavior in the evalutation of the operands on +=. As for (3), the operation of the compound assignment += (more precisely the side effect of that operation; its value computation, if needed, is in any case sequenced after its side effect) now must both fetch the current value of i, and then (obviously sequenced after it, even if the standard does not say so explicitly, or otherwise the evaluation of such operators would always invoke undefined behavior) add the RHS and store the result back into i. Both these operations would have given undefined behavior if they were unsequenced w.r.t. the side effect of the ++, but as argued above (the side effect of the ++ is sequenced before the value computation of + giving the RHS of the += operator, which value computation is sequenced before the operation of that compound assignment), that is not the case.
But on the other hand the standard also says that E += F is equivalent to E = E + F, except that (the lvalue) E is evaluated only once. Now in our example the value computation of i (which is what E is here) as lvalue does not involve anything that needs to be sequenced w.r.t. other actions, so doing it once or twice makes no difference; our expression should be strictly equivalent to E = E + F. But here's the problem; it is pretty obvious that evaluating i = i + (++i + 1) would give undefined behaviour! What gives? Or is this a defect of the standard?
Added. I have slightly modified my discussion above, to do more justice to the proper distinction between side effects and value computations, and using "evaluation" (as does the standard) of an expression to encompass both. I think my main interrogation is not just about whether behavior is defined or not in this example, but how one must read the standard in order to decide this. Notably, should one take the equivalence of E op= F to E = E op F as the ultimate authority for the semantics of the compound assignment operation (in which case the example clearly has UB), or merely as an indication of what mathematical operation is involved in determining the value to be assigned (namely the one identified by op, with the lvalue-to-rvalue converted LHS of the compound assignment operator as left operand and its RHS as right operand). The latter option makes it much harder to argue for UB in this example, as I have tried to explain. I admit that it is tempting to make the equivalence authoritative (so that compound assignments become a kind of second-class primitives, whose meaning is given by rewriting in term of first-class primitives; thus the language definition would be simplified), but there are rather strong arguments against this:
The equivalence is not absolute, because of the "E is evaluated only once" exception. Note that this exception is essential to avoid making any use where the evaluation of E involves a side effect undefined behavior, for instance in the fairly common a[i++] += b; usage. If fact I think no absolutely equivalent rewriting to eliminate compound assignments is possible; using a fictive ||| operator to designate unsequenced evaluations, one might try to define E op= F; (with int operands for simplicity) as equivalent to { int& L=E ||| int R=F; L = L + R; }, but then the example no longer has UB. In any case the standard gives us no rewriitng recipe.
The standard does not treat compound assignments as second-class primitives for which no separate definition of semantics is necessary. For instance in 5.17 (emphasis mine)
The assignment operator (=) and the compound assignment operators all group right-to-left. [...] In all cases, the assignment is sequenced after the value
computation of the right and left operands, and before the value computation of the assignment expression. With respect to an indeterminately-sequenced function call, the operation of a compound assignment is a single evaluation.
If the intention were to let compound assignments be mere shorthands for simple assignments, there would be no reason to include them explicitly in this description. The final phrase even directly contradicts what would be the case if the equivalence was taken to be authoritative.
If one admits that compound assignments have a semantics of their own, then the point arises that their evaluation involves (apart from the mathematical operation) more than just a side effect (the assignment) and a value evaluation (sequenced after the assignment), but also an unnamed operation of fetching the (previous) value of the LHS. This would normally be dealt with under the heading of "lvalue-to-rvalue conversion", but doing so here is hard to justify, since there is no operator present that takes the LHS as an rvalue operand (though there is one in the expanded "equivalent" form). It is precisely this unnamed operation whose potential unsequenced relation with the side effect of ++ would cause UB, but this unsequenced relation is nowhere explicitly stated in the standard, because the unnamed operation is not. It is hard to justify UB using an operation whose very existence is only implicit in the standard.
About the description of i = ++i + 1
I gather that the subtle explanation is that
(1) the expression ++i returns an lvalue but + takes prvalues as operands, so a conversion from lvalue to prvalue must be performed;
Probably, see CWG active issue 1642.
this involves obtaining the
current value of that lvalue (rather than one more than the old value
of i) and must therefore be sequenced after the side effect from the
increment (i.e., updating i)
The sequencing here is defined for the increment (indirectly, via +=, see (a)):
The side effect of ++ (the modification of i) is sequenced before the value computation of the whole expression ++i. The latter refers to computing the result of ++i, not to loading the value of i.
(2) the LHS of the assignment is also an
lvalue, so its value evaluation does not involve fetching the current
value of i;
while this value computation is unsequenced w.r.t. the
value computation of the RHS, this poses no problem
I don't think that's properly defined in the Standard, but I'd agree.
(3) the value
computation of the assignment itself involves updating i (again),
The value computation of i = expr is only required when you use the result, e.g. int x = (i = expr); or (i = expr) = 42;. The value computation itself does not modify i.
The modification of i in the expression i = expr that happens because of the = is called the side effect of =. This side effect is sequenced before value computation of i = expr -- or rather the value computation of i = expr is sequenced after the side effect of the assignment in i = expr.
In general, the value computation of the operands of an expression are sequenced before the side effect of that expression, of course.
but is sequenced after the value computation of its RHS, and hence after
the previous update to i; no problem.
The side effect of the assignment i = expr is sequenced after the value computation of the operands i (A) and expr of the assignment.
The expr in this case is a +-expression: expr1 + 1. The value computation of this expression is sequenced after the value computations of its operands expr1 and 1.
The expr1 here is ++i. The value computation of ++i is sequenced after the side effect of ++i (the modification of i) (B)
That's why i = ++i + 1 is safe: There's a chain of sequenced before between the value computation in (A) and the side effect on the same variable in (B).
(a) The Standard defines ++expr in terms of expr += 1, which is defined as expr = expr + 1 with expr being evaluated only once.
For this expr = expr + 1, we therefore have only one value computation of expr. The side effect of = is sequenced before the value computation of the whole expr = expr + 1, and it's sequenced after the value computation of the operands expr (LHS) and expr + 1 (RHS).
This corresponds to my claim that for ++expr, the side effect is sequenced before the value computation of ++expr.
About i += ++i + 1
Does the value computation of i += ++i + 1 involve undefined behavior?
Since the
LHS of += is still an lvalue (and its RHS still a prvalue), the same
reasoning as above applies as far as (1) and (2) are concerned;
as for
(3) the value computation of the += operator now must both fetch the
current value of i, and then (obviously sequenced after it, even if
the standard does not say so explicitly, or otherwise the execution of
such operators would always invoke undefined behavior) perform the
addition of the RHS and store the result back into i.
I think here's the problem: The addition of i in the LHS of i += to the result of ++i + 1 requires knowing the value of i - a value computation (which can mean loading the value of i). This value computation is unsequenced with respect to the modification performed by ++i. This is essentially what you say in your alternative description, following the rewrite mandated by the Standard i += expr -> i = i + expr. Here, the value computation of i within i + expr is unsequenced with respect to the value computation of expr. That's where you get UB.
Please note that a value computation can have two results: The "address" of an object, or the value of an object. In an expression i = 42, the value computation of the lhs "produces the address" of i; that is, the compiler needs to figure out where to store the rhs (under the rules of observable behaviour of the abstract machine). In an expression i + 42, the value computation of i produces the value. In the above paragraph, I was referring to the second kind, hence [intro.execution]p15 applies:
If a side effect on a scalar object is unsequenced relative to either
another side effect on the same scalar object or a value computation
using the value of the same scalar object, the behavior is undefined.
Another approach for i += ++i + 1
the value computation of the += operator now must both fetch the
current value of i, and then [...] perform the addition of the RHS
The RHS being ++i + 1. Computing the result of this expression (the value computation) is unsequenced with respect to the value computation of i from the LHS. So the word then in this sentence is misleading: Of course, it must first load i and then add the result of the RHS to it. But there's no order between the side-effect of the RHS and the value computation to get the value of the LHS. For example, you could get for the LHS either the old or the new value of i, as modified by the RHS.
In general a store and a "concurrent" load is a data race, which leads to Undefined Behaviour.
Addressing the addendum
using a fictive ||| operator to designate unsequenced evaluations, one might try to define E op= F; (with int operands for simplicity) as equivalent to { int& L=E ||| int R=F; L = L + R; }, but then the example no longer has UB.
Let E be i and F be ++i (we don't need the + 1). Then, for i = ++i
int* lhs_address;
int lhs_value;
int* rhs_address;
int rhs_value;
( lhs_address = &i)
||| (i = i+1, rhs_address = &i, rhs_value = *rhs_address);
*lhs_address = rhs_value;
On the other hand, for i += ++i
( lhs_address = &i, lhs_value = *lhs_address)
||| (i = i+1, rhs_address = &i, rhs_value = *rhs_address);
int total_value = lhs_value + rhs_value;
*lhs_address = total_value;
This is intended to represent my understanding of the sequencing guarantees. Note that the , operator sequences all value computations and side effects of the LHS before those of the RHS. Parentheses do not affect sequencing. In the second case, i += ++i, we have a modification of i unsequenced wrt an lvalue-to-rvalue conversion of i => UB.
The standard does not treat compound assignments as second-class primitives for which no separate definition of semantics is necessary.
I would say that's a redundancy. The rewrite from E1 op = E2 to E1 = E1 op E2 also includes which expression types and value categories are required (on the rhs, 5.17/1 says something about the lhs), what happens to pointer types, the required conversions etc. The sad thing is that the sentence about "With respect to an.." in 5.17/1 is not in 5.17/7 as an exception of that equivalence.
In any way, I think we should compare the guarantees and requirements for compound assignment vs. simple assignment plus the operator, and see if there's any contradiction.
Once we put that "With respect to an.." also in the list of exceptions in 5.17/7, I don't think there's a contradiction.
As it turns out, as you can see in the discussion of Marc van Leeuwen's answer, this sentence leads to the following interesting observation:
int i; // global
int& f() { return ++i; }
int main() {
i = i + f(); // (A)
i += f(); // (B)
}
It seems that (A) has an two possible outcomes, since the evaluation of the body of f is indeterminately sequenced with the value computation of the i in i + f().
In (B), on the other hand, the evaluation of the body of f() is sequenced before the value computation of i, since += must be seen as a single operation, and f() certainly needs to be evaluated before the assignment of +=.
The expression:
i += ++i + 1
does invoke undefined behavior. The language lawyer method requires us to go back to the defect report that results in:
i = ++i + 1 ;
becoming well defined in C++11, which is defect report 637. Sequencing rules and example disagree , it starts out saying:
In 1.9 [intro.execution] paragraph 16, the following expression is
still listed as an example of undefined behavior:
i = ++i + 1;
However, it appears that the new sequencing rules make this expression
well-defined
The logic used in the report is as follows:
The assignment side-effect is required to be sequenced after the value computations of both its LHS and RHS (5.17 [expr.ass] paragraph 1).
The LHS (i) is an lvalue, so its value computation involves computing the address of i.
In order to value-compute the RHS (++i + 1), it is necessary to first value-compute the lvalue expression ++i and then do an lvalue-to-rvalue conversion on the result. This guarantees that the incrementation side-effect is sequenced before the computation of the addition operation, which in turn is sequenced before the assignment side effect. In other words, it yields a well-defined order and final value for this expression.
So in this question our problem changes the RHS which goes from:
++i + 1
to:
i + ++i + 1
due to draft C++11 standard section 5.17 Assignment and compound assignment operators which says:
The behavior of an expression of the form E1 op = E2 is equivalent to
E1 = E1 op E2 except that E1 is evaluated only once. [...]
So now we have a situation where the computation of i in the RHS is not sequenced relative to ++i and so we then have undefined behavior. This follows from section 1.9 paragraph 15 which says:
Except where noted, evaluations of operands of individual operators
and of subexpressions of individual expressions are unsequenced. [
Note: In an expression that is evaluated more than once during the
execution of a program, unsequenced and indeterminately sequenced
evaluations of its subexpressions need not be performed consistently
in different evaluations. —end note ] The value computations of the
operands of an operator are sequenced before the value computation of
the result of the operator. If a side effect on a scalar object is
unsequenced relative to either another side effect on the same scalar
object or a value computation using the value of the same scalar
object, the behavior is undefined.
The pragmatic way to show this would be to use clang to test the code, which generates the following warning (see it live):
warning: unsequenced modification and access to 'i' [-Wunsequenced]
i += ++i + 1 ;
~~ ^
for this code:
int main()
{
int i = 0 ;
i += ++i + 1 ;
}
This is further bolstered by this explicit test example in clang's test suite for -Wunsequenced:
a += ++a;
Yes, it is UB!
The evaluation of your expression
i += ++i + 1
proceeds in the following steps:
5.17p1 (C++11) states (emphases mine):
The assignment operator (=) and the compound assignment operators all group right-to-left. All require a modifiable lvalue as their left operand and return an lvalue referring to the left operand. The result in all cases is a bit-field if the left operand is a bit-field. In all cases, the assignment is sequenced after the value computation of the right and left operands, and before the value computation of the assignment expression.
What does "value computation" mean?
1.9p12 gives the answer:
Accessing an object designated by a volatile glvalue (3.10), modifying an object, calling a library I/O function, or calling a function that does any of those operations are all side effects, which are changes in the state of the execution environment. Evaluation of an expression (or a sub-expression) in general includes both value computations (including determining the identity of an object for glvalue evaluation and fetching a value previously assigned to an object for prvalue evaluation) and initiation of side effects.
Since your code uses a compound assignment operator, 5.17p7 tells us, how this operator behaves:
The behavior of an expression of the form E1 op= E2 is equivalent to E1 = E1 op E2 except that E1 is evaluated only once.
Hence the evaluation of the expression E1 ( == i) involves both, determining the identity of the object designated by i and an lvalue-to-rvalue conversion to fetch the value stored in that object. But the evaluation of the two operands E1 and E2 are not sequenced with respect to each other. Thus we get undefined behavior since the evaluation of E2 ( == ++i + 1) initiates a side effect (updating i).
1.9p15:
... If a side effect on a scalar object is unsequenced relative to either another side effect on the same scalar object or a value computation using the value of the same scalar object, the behavior is undefined.
The following statements in your question/comments seem to be the root of your misunderstanding:
(2) the LHS of the assignment is also an lvalue, so its value evaluation does not involve fetching the current value of i
fetching a value can be part of a prvalue evaluation. But in E += F the only prvalue is F so fetching the value of E is not part of the evaluation of the (lvalue) subexpression E
If an expression is an lvalue or rvalue doesn't tell anything about how this expression is to be evaluated. Some operators require lvalues as their operands some others require rvalues.
Clause 5p8:
Whenever a glvalue expression appears as an operand of an operator that expects a prvalue for that operand, the lvalue-to-rvalue (4.1), array-to-pointer (4.2), or function-to-pointer (4.3) standard conversions are applied to convert the expression to a prvalue.
In a simple assignment the evaluation of of the LHS only requires determining the identity of the object. But in a compound assignment such as += the LHS must be a modifiable lvalue, but the evaluation of the LHS in this case consists of determining the identity of the object and an lvalue-to-rvalue conversion. It is the result of this conversion (which is a prvalue) that is added to the result (also a prvalue) of the evaluation of the RHS.
"But in E += F the only prvalue is F so fetching the value of E is not part of the evaluation of the (lvalue) subexpression E"
That's not true as I explained above. In your example F is a prvalue expression, but F may as well be an lvalue expression. In that case, the lvalue-to-rvalue conversion is also applied to F. 5.17p7 as cited above tells us, what the semantics of the compound assignment operators are. The standard states that the behavior of E += F is the same as of E = E + F but E is only evaluated once. Here, the evaluation of E includes the lvalue-to-rvalue conversion, because the binary operator + requires it operands to be rvalues.
There is no clear case for Undefined Behavior here
Sure, an argument leading to UB can be given, as I indicated in the question, and which has been repeated in the answers given so far. However this involves a strict reading of 5.17:7 that is both self-contradictory and in contradiction with explicit statements in 5.17:1 about compound assignment. With a weaker reading of 5.17:7 the contradictions disappear, as does the argument for UB. Whence my conclusion is neither that there is UB here, nor that there is clearly defined behaviour, but the the text of the standard is inconsistent, and should be modified to make clear which reading prevails (and I suppose this means a defect report should be written). Of course one might invoke here the fall-back clause in the standard (the note in 1.3.24) that evaluations for which the standard fails to define the behavior [unambiguously and self-consistently] are Undefined Behavior, but that would make any use of compound assignments (including prefix increment/decrement operators) into UB, something that might appeal to certain implementors, but certainly not to programmers.
Instead of arguing for the given problem, let me present a slightly modified example that brings out the inconsistency more clearly. Assume one has defined
int& f (int& a) { return a; }
a function that does nothing and returns its (lvalue) argument. Now modify the example to
n += f(++n) + 1;
Note that while some extra conditions about sequencing of function calls are given in the standard, this would at first glance not seem to effect the example, since there are no side effect at all from the function call (not even locally inside the function), as the incrementation happens in the argument expression for f, whose evaluation is not subject to those extra conditions. Indeed, let us apply the Crucial Argument for Undefined Behavior (CAUB), namely 5.17:7 which says that the behavior of such a compound assignment is equivalent to that of (in this case)
n = n + f(++n) + 1;
except that n is evaluated only once (an exception that makes no difference here). The evaluation of the statement I just wrote clearly has UB (the value computation of the first (prvalue) n in the RHS is unsequenced w.r.t. the side effect of the ++ operation, which involves the same scalar object (1.9:15) and you're dead).
So the evaluation of n += f(++n) + 1 has undefined behavior, right? Wrong! Read in 5.17:1 that
With respect to an indeterminately-sequenced function call, the operation of a compound assignment is a single evaluation. [ Note: Therefore, a function call shall not intervene between the lvalue-to-rvalue conversion and the side effect associated with any single compound assignment operator. — end note ]
This language is far from as precise as I would like it to be, but I don't think it is a stretch to assume that "indeterminately-sequenced" should mean "with respect to that operation of a compound assignment". The (non normative, I know) note makes it clear that the lvalue-to-rvalue conversion is part of the operation of the compound assignment. Now is the call of f indeterminately-sequenced with respect to the operation of the compound assignment of +=? I'm unsure, because the 'sequenced' relation is defined for individual value computations and side effects, not complete evaluations of operators, which may involve both. In fact the evaluation of a compound assignment operator involves three items: the lvalue-to-rvalue conversion of its left operand, the side effect (the assignment proper), and the value computation of the compound assignment (which is sequenced after the side effect, and returns the original left operand as lvalue). Note that the existence of the lvalue-to-rvalue conversion is never explicitly mentioned in the standard except in the note cited above; in particular, the standard makes no (other) statement at all regarding its sequencing relative to other evaluations. It is pretty clear that in the example the call of f is sequenced before the side effect and value computation of += (since the call occurs in the value computation of the right operand to +=), but it might be indeterminately-sequenced with respect to the lvalue-to-rvalue conversion part. I recall from my question that since the left operand of += is an lvalue (and necessarily so), one cannot construe the lvalue-to-rvalue conversion to have occurred as part of the value computation of the left operand.
However, by the principle of the excluded middle, the call to f must either be indeterminately-sequenced with respect to the operation of the compound assignment of +=, or not indeterminately-sequenced with respect to it; in the latter case it must be sequenced before it because it cannot possibly be sequenced after it (the call of f being sequenced before the side effect of +=, and the relation being anti-symmetric). So first assume it is indeterminately-sequenced with respect to the operation. Then the cited clause says that w.r.t. the call of f the evaluation of += is a single operation, and the note explains that it means the call should not intervene between the lvalue-to-rvalue conversion and the side effect associated with +=; it should either be sequenced before both, or after both. But being sequenced after the side effect is not possible, so it should be before both. This makes (by transitivity) the side effect of ++ sequenced before the lvalue-to-rvalue conversion, exit UB. Next assume the call of f is sequenced before the operation of +=. Then it is in particular sequenced before the lvalue-to-rvalue conversion, and again by transitivity so is the side effect of ++; no UB in this branch either.
Conclusion: 5.17:1 contradicts 5.17:7 if the latter is taken (CAUB) to be normative for questions of UB resulting from unsequenced evaluations by 1.9:15. As I said CAUB is self-contradictory as well (by arguments indicated in the question), but this answer is getting to long, so I'll leave it at this for now.
Three problems, and two proposals for resolving them
Trying to understand what the standard writes about these matters, I distinguish three aspects in which the text is hard to interpret; they all are of a nature that the text is insufficiently clear about what model its statements are referring to. (I cite the texts at the end of the numbered items, since I do not know the markup to resume a numbered item after a quote)
The text of 5.17:7 is of an apparent simplicity that, although the intention is easy to grasp, gives us little hold when applied to difficult situations. It makes a sweeping claim (equivalent behavior, apparently in all aspects) but whose application is thwarted by the exception clause. What if the behavior of E1 = E1 op E2 is undefined? Well then that of E1 op = E2 should be as well. But what if the UB was due to E1 being evaluated twice in E1 = E1 op E2? Then evaluating E1 op = E2 should presumably not be UB, but if so, then defined as what? This is like saying "the youth of the second twin was exactly like that of the first, except that he did not die at childbirth." Frankly, I think this text, which has little evolved since the C version "A compound assignment of the the form E1 op = E2 differs from the simple assignment expression E1 = E1 op E2 only in that the lvalue E1 is evaluated only once." might be adapted to match the changes in the standard.
(5.17) 7 The behavior of an expression of the form E1 op = E2 is equivalent to
E1 = E1 op E2 except that E1 is evaluated only once.[...]
It is not so clear what precisely the actions (evaluations) are between which the 'sequenced' relation is defined. It is said (1.9:12) that evaluation of an expression includes value computations and initiation of side effects. Though this appears to say that an evaluation may have multiple (atomic) components, the sequenced relation is actually mostly defined (e.g. in 1.9:14,15) for individual components, so that it might be better to read this as that the notion of "evaluation" encompasses both value computations and (initiation of) side effects. However in some cases the 'sequenced' relation is defined for the (entire) execution of an expression of statement (1.9:15) or for a function call (5.17:1), even though a passage in 1.9:15 avoids the latter by referring directly to executions in the body of a called function.
(1.9) 12 Evaluation of an expression (or a sub-expression) in general includes
both value computations (...) and initiation of side effects. [...] 13 Sequenced before is an asymmetric, transitive, pair-wise relation between evaluations executed by a single thread [...] 14 Every value computation and side effect associated with a full-expression is sequenced before every value computation and side effect associated with the next full-expression to be evaluated. [...] 15 When calling a function (whether or not the function is inline), every value computation and side effect
associated with any argument expression, or with the postfix expression designating the called function, is
sequenced before execution of every expression or statement in the body of the called function. [...] Every evaluation in the calling function (including other function calls) ... is indeterminately sequenced with
respect to the execution of the called function [...] (5.2.6, 5.17) 1 ... With respect to an indeterminately-sequenced function call, ...
The text should more clearly acknowledge that a compound assignment involves, in contrast to a simple assignment, the action of fetching the value previously assigned to its left operand; this action is like lvalue-to-rvalue conversion, but does not happen as part of the value computation of that left operand, since it is not a prvalue; indeed it is a problem that 1.9:12 only acknowledges such action for prvalue evaluation. In particular the text should be more clear about which 'sequenced' relations are given for that action, if any.
(1.9) 12 Evaluation of an expression... includes... value computations (including determining the identity of an object for glvalue evaluation and fetching a value previously assigned to an object for prvalue evaluation)
The second point is the least directly related to our concrete question, and I think it can be solved simply by choosing a clear point of view and reformulating pasages that seem to indicate a different point of view. Given that one of the main purposes of the old sequence points, and now the 'sequenced' relation, was to make clear that the side effect of postfix-increment operators is unsequenced w.r.t. to actions sequenced after the value computation of that operator (thus giving e.g. i = i++ UB), the point of view must be that individual value computations and (initiation of) individual side effects are "evaluations" for which "sequenced before" may be defined. For pragmatic reasons I would also include two more kinds of (trivial) "evaluations": function entry (so that the language of 1.9:15 may be simplified to: "When calling a function..., every value computation and side effect associated with any of its argument expressions, or with the postfix expression designating the called function, is sequenced before entry of that function") and function exit (so that any action in the function body gets by transitivity sequenced before anything that requires the function value; this used to be guaranteed by a sequence point, but the C++11 standard seems to have lost such guarantee; this might make calling a function ending with return i++; potentially UB where this is not intended, and used to be safe). Then one can also be clear about the "indeterminately sequenced" relation of functions calls: for every function call, and every evaluation that is not (directly or indirectly) part of evaluating that call, that evaluation shall be sequenced (either before or after) w.r.t. both entry and exit of that function call, and it shall have the same relation in both cases (so that in particular such external actions cannot be sequenced after function entry but before function exit, as is clearly desirable within a single thread).
Now to resolve points 1. and 3., I can see two paths (each affecting both points), which have different consequences for the defined or not behavior of our example:
Compound assignments with two operands, and three evaluations
Compound operations have thier two usual operands, an lvalue left operand and a prvalue right operand. To settle the unclarity of 3., it is included in 1.9:12 that fetching the value previously assigned to an object also may occur in compound assignments (rather than only for prvalue evaluation). The semantics of compount assignments are defined by changing 5.17:7 to
In a compound assignment op=, the value previously assigned to the object referred to by the left operand is fetched, the operator op is applied with this value as left operand and the right operand of op= as right operand, and the resulting value replaces that of the object referred to by the left operand.
(That gives two evaluations, the fetch and the side effect; a third evaluation is the trivial value computation of the compound operator, sequenced after both other evaluations.)
For clarity, state clearly in 1.9:15 that value computations in operands are sequenced before all value computations associated with the operator (rather than just those for the result of the operator), which ensures that evaluating the lvalue left operand is sequenced before fetching its value (one can hardly imagine otherwise), and also sequences the value computation of the right operand before that fetch, thus excluding UB in our example. While at it, I see no reason not to also sequence value computations in operands before any side effects associated with the operator (as they clearly must); this would make mentioning this explicitly for (compound) assignments in 5.17:1 superfluous. On the other hand do mention there that the value fetching in a compound assignment is sequenced before its side effect.
Compound assignments with three operands, and two evaluations
In order to obtain that the fetch in a compount assignment will be unsequenced with respect to the value computation of the right operand, making our example UB, the clearest way seems to be to give compound operators an implicit third (middle) operand, a prvalue, not represented by a separate expression, but obtained by lvalue-to-rvalue conversion from the left operand (this three-operand nature corresponds to the expanded form of compound assignments, but by obtaining the middle operand from the left operand, it is ensured that the value is fetched from the same object to which the result will be stored, a crucial guarantee that is only vaguely and implicitly given in the current formulation through the "except that E1 is evaluated only once" clause). The difference with the previous solution is that the fetch is now a genuine lvalue-to-rvalue conversion (since the middle operand is a prvalue) and is performed as part of the value computation of the operands to the compound assignment, which makes it naturally unsequenced with the value computation of the right operand. It should be stated somewhere (in a new clause that describes this implicit operand) that the value computation of the left operand is sequenced before this lvalue-to-rvalue conversion (it clearly must). Now 1.9:12 can be left as it is, and in place of 5.17:7 I propose
In a compound assignment op= with left operand a (an lvalue), and midlle and right operands brespectively c (both prvalues), the operator op is applied with b as left operand and c as right operand, and the resulting value replaces that of the object referred to by a.
(That gives one evaluation, the side effect, with as second evaluation the trivial value computation of the compound operator, sequenced after it.)
The still applicable changes to 1.9:15 and 5.17:1 suggested in the previous solution could still apply, but would not give our original example defined behavior. However the modified example at the top of this answer would still have defined behavior, unless the part 5.17:1 "compound assignment is a single operation" is scrapped or modified (there is a similar passage in 5.2.6 for postfix increment/decrement). The existence of those passages would suggest that detaching the fecth and store operations within a single compound assignement or postfix increment/decrement was not the intention of those who wrote the current standard (and by extension making our example UB), but this of course is mere guesswork.
From the compiler writer's perspective, they don't care about "i += ++i + 1", because whatever the compiler does, the programmer may not get the correct result, but they surely get what they deserve. And nobody writes code like that. What the compiler writer cares about is
*p += ++(*q) + 1;
The code must read *p and *q, increase *q by 1, and increase *p by some amount that is calculated. Here the compiler writer cares about restrictions on the order of read and write operations. Obviously if p and q point to different objects, the order makes no difference, but if p == q then it will make a difference. Again, p will be different from q unless the programmer writing the code is insane.
By making the code undefined, the language allows the compiler to produce the fastest possible code without caring for insane programmers. By making the code defined, the language forces the compiler to produce code that conforms to the standard even in insane cases, which may make it run slower. Both compiler writers and sane programmers don't like that.
So even if the behaviour is defined in C++11, it would be very dangerous to use it, because (a) a compiler might not be changed from C++03 behaviour, and (b) it might be undefined behaviour in C++14, for the reasons above.
The C++11 standard (5.17, expr.ass) states that
In all cases, the assignment is sequenced after the value computation
of the right and left operands, and before the value computation of
the assignment expression. With respect to an
indeterminately-sequenced function call, the operation of a compound
assignment is a single evaluation
As I understand it, all expressions which are a part of the given assignment will be evaluated before the assignment itself. This rule should work even if I modify the same variable twice in the same assignment, which, I am fairly certain, was undefined behavior before.
Will the given code:
int a = 0;
a = (a+=1) = 10;
if ( a == 10 ) {
printf("this is defined");
} else {
printf("undefined");
}
always evaluate to a==10?
Yes, there was a change between C++98 and C++11. I believe your example to be well-defined under C++11 rules, while exhibiting undefined behavior under C++98 rules.
As a simpler example, x = ++x; is undefined in C++98 but is well-defined in C++11. Note that x = x++; is still undefined (side effect of post-increment is unsequenced with the evaluation of the expression, while side effect of pre-increment is sequenced before the same).
Let's rewrite your code as
E1 = (E2 = E3)
where E1 is the expression a, E2 is the expression a += 1 and E3 is the expression 10. Here we ussed, that the assignment operator groups right-to-left (§5.17/1 in C++11 Standard).
§5.17/1 moreover states:
In all cases, the assignment is sequenced after the value computation of the right and left operands, and before the value computation of the assignment expression.
Applying this to our expression means that we first must evaluate the subexpressions E1 and E2 = E3. Note that there is no "sequenced-before" relationship between these two evaluations, but that causes no problems.
The evaluation of the id-expression E1 is trivial (the result is a itself). The evaluation of the assignment-expression E2 = E3 proceeds as follows:
First both subexpressions have to be evaluated. The evaluation of the literal E3is again trivial (gives a prvalue of value 10).
The evaluation of the (compound) assignment-expression E2 is done in the following steps:
1) The behavior of a += 1is equivalent to a = a + 1 but a is only evaluated once (§5.17/7). After evaluating the subexpressions a and 1 (in an arbitrary order), an lvalue-to-rvalue conversion is applied to a in order to read the value stored in a.
2) The values of a (which is 0) and of 1 are added (a + 1) and the result of this addition is a prvalue of value 1.
3) Before we can compute the result of the assignment a = a + 1 the value of the object the left operand refers to is replaced by the value of the right operand (§5.17/2). The result of E2 is then an lvalue refereing to the new value 1. Note that the side effect (updating the value of the left operand) is sequenced before the value computation of the assignment expression. This is §5.17/1 cited above.
Now that we have evaluated the subexpressions E2and E3, the value of the expression E2refers to is replaced by the value of E3, which is 10. Hence the result of E2 = E3 is an lvalue of value 10.
Finally, the value expression E1 refers to is replaced by the value of the expression E2 = E3, which we computed to be 10. Thus, the variable aends up to contain the value 10.
Since all these steps are well-defined, the whole expression yields a well-defined value.
After doing a little research, I am convinced your codes behaviour is well defined in C++11.
$1.9/15 states:
The value computations of the operands of an operator are sequenced before
the value computation of the result of the operator.
$5.17/1 states:
The assignment operator (=) and the compound assignment operators all group
right-to-left.
If I understand correctly, in your example
a = (a+=1) = 10;
this implies that the value computations of (a+=1) and 10 have to be made before the value computation of (a+=1) = 10 and the value computation of this expression has to be finished before a = (a+=1) = 10; is evaluated.
$5.17/1 states:
In all cases, the assignment is sequenced after the value computation of the right and left operands, and before the value computation of the assignment expression.
This implies that the assignment must happen before the value computation, and therefore, due to transitivity, the evaluation of (a+=1) = 10 can only begin after the assignment a+=1 (Because its value may only be computed after the side effect).
The same is true for the second and third assignment.
See also this excellent answer, which explains the sequenced-before relation in much more detail and way better than I could.