Related
sorry if my question is duplicate or not worth answering.
I have been the following code that produces the result that I am not understanding how.
int x=5;
int y;
y = ++x * ++x;
cout<<x <<endl;
cout<<y;
According to my little understanding of programming, the value of y should be 42, but, the output on the computer is 49. Kindly help what will be the output of variable y.
I am executing the code in Dev-C++.
Thanks in advance.
The short answer is that you have an undefined behavior in there.
The exact answer depends on which standard you use.
Pre C++11, we had the notion of sequence points. Intuitively, they are points in which all the values have been properly stored, such as at the end of statement (at the semicolon). The standard says that
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.
, which means that between 2 sequence points (for the sake of simplicity, read as between 2 semicolons), the value of a variable cannot be changed more than once. You change the value twice, using the increment operator.
C++11 removed the notion of sequence points with relationships of sequenced before, sequenced after or unsequenced, referring to the order in which the expressions are evaluated.
In arithmetic expressions,
If a side effect on a scalar object is unsequenced relative to either another side effect on the same scalar object or [...] the behaviour is undefined.
and there is no sequencing between the operators of an arithmetic expression. Therefore, it is still a case of undefined behavior, just for another reason.
This means in practice that the compiler can choose what to do, and, in your case, it produces the results you observe. You should try to avoid undefined behaviour in your programs as much as possible. Below are some references that expand on the subject:
https://en.cppreference.com/w/cpp/language/eval_order
Undefined behavior and sequence points
vec[ival++] <= vec[ival]
This expression has undefined behavior, because the order of evaluation of operator's (<=) operands is undefined.
How can we rewrite that expression to avoid the undefined behavior?
I've found an answer that appears to work:
vec[ival] <= vec[ival + 1]
If that is the right way to avoid the undefined behavior, why does rewriting it that way avoid the undefined behavior?
Adding any reference about how to fix that expression would be great.
Yes, your first example has undefined behavior because we have an unsequenced modification and access of a memory location, which is undefined behavior. This is covered in draft C++ standard [intro.execution]p10 (emphasis mine):
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 memory location
([intro.memory]) is unsequenced relative to either another side effect
on the same memory location or a value computation using the value of
any object in the same memory location, and they are not potentially
concurrent ([intro.multithread]), the behavior is undefined. [ Note:
The next subclause imposes similar, but more complex restrictions on
potentially concurrent computations. — end note ] [ Example:
void g(int i) {
i = 7, i++, i++; // i becomes 9
i = i++ + 1; // the value of i is incremented
i = i++ + i; // the behavior is undefined
i = i + 1; // the value of i is incremented
}
— end example ]
If we check out the section of relational operators which covers <= [expr.rel] it does not specify an order of evaluation, so we are covered by intro.exection and thus we have undefined behavior.
Having unspecified order of evaluation is not sufficient for undefined behavior as the example in Order of evaluation of assignment statement in C++ demonstrates.
Your second example avoids the undefined behavior since you are not introducing a side effect to ival, you are just reading the memory location twice.
I believe that is a reasonable way to solve the problem, it is readable and not surprising. An alternate method could include introducing a second variable, e.g. index and prev_index. It is hard to come with a fast rule given such a small code snippet.
It avoids undefined behavior because you are not changing the value of ival. The issue you're seeing in the first sample is that we can't determine what the values of ival are at the times that they're used. In the second sample, there's no confusion.
Let's start with the worst problem first, and that is the Undefined Behavior. C++ uses sequencing rules. Statements are executed in seqeuence. Usually that's top-to-bottom, but if statements, for statements, function calls and the like can change that.
Now withing a statement there still might be a further sequence of execution, but I'm very intentionally writing might. By defaults, the various parts of a single statement are not sequenced relative to each other. That's why you can get the varying order of execution. But worse, if you change and use a single object without sequencing, you have Undefined Behavior. That is bad - anything might happen.
The proposed solution (iVal + 1) doesn't change iVal anymore, but generates a new value. That is entirely safe. Still, it leaves iVal unchanged.
You may want to check on std::adjacent_find(). Chances are that the loop you're trying to write is already in the Standard Library.
The first problem is that as the initial code exhibited undefined behavior, under the C++ standard there is no "fix". The behavior of that line of code is not specified by the C++ standard; to know what it is supposed to do you have to have another source of information.
Formally, that expression can be rewritten as system("format c:"), as the C++ standard does not mandate any behavior from a program that exhibits undefined behavior.
But in practice, when you run into something like that, you have to read the original programmer's mind.
Well, you can solve anything with lambdas.
[&]{ bool ret = vec[ival] <= vec[ival+1]; ++ival; return ret; }()
Second,
vec[ival] <= vec[ival+1]
is not the same, because it lacks the side effect of ival being 1 greater after the expression is evaluated.
I was reading about order of evaluation violations, and they give an example that puzzles me.
1) If a side effect on a scalar object is un-sequenced relative to another side effect on the same scalar object, the behavior is undefined.
// snip
f(i = -1, i = -1); // undefined behavior
In this context, i is a scalar object, which apparently means
Arithmetic types (3.9.1), enumeration types, pointer types, pointer to member types (3.9.2), std::nullptr_t, and cv-qualified versions of these types (3.9.3) are collectively called scalar types.
I don’t see how the statement is ambiguous in that case. It seems to me that regardless of if the first or second argument is evaluated first, i ends up as -1, and both arguments are also -1.
Can someone please clarify?
UPDATE
I really appreciate all the discussion. So far, I like #harmic’s answer a lot since it exposes the pitfalls and intricacies of defining this statement in spite of how straight forward it looks at first glance. #acheong87 points out some issues that come up when using references, but I think that's orthogonal to the unsequenced side effects aspect of this question.
SUMMARY
Since this question got a ton of attention, I will summarize the main points/answers. First, allow me a small digression to point out that "why" can have closely related yet subtly different meanings, namely "for what cause", "for what reason", and "for what purpose". I will group the answers by which of those meanings of "why" they addressed.
for what cause
The main answer here comes from Paul Draper, with Martin J contributing a similar but not as extensive answer. Paul Draper's answer boils down to
It is undefined behavior because it is not defined what the behavior is.
The answer is overall very good in terms of explaining what the C++ standard says. It also addresses some related cases of UB such as f(++i, ++i); and f(i=1, i=-1);. In the first of the related cases, it's not clear if the first argument should be i+1 and the second i+2 or vice versa; in the second, it's not clear if i should be 1 or -1 after the function call. Both of these cases are UB because they fall under the following rule:
If a side effect on a scalar object is unsequenced relative to another side effect on the same scalar object, the behavior is undefined.
Therefore, f(i=-1, i=-1) is also UB since it falls under the same rule, despite that the intention of the programmer is (IMHO) obvious and unambiguous.
Paul Draper also makes it explicit in his conclusion that
Could it have been defined behavior? Yes. Was it defined? No.
which brings us to the question of "for what reason/purpose was f(i=-1, i=-1) left as undefined behavior?"
for what reason / purpose
Although there are some oversights (maybe careless) in the C++ standard, many omissions are well-reasoned and serve a specific purpose. Although I am aware that the purpose is often either "make the compiler-writer's job easier", or "faster code", I was mainly interested to know if there is a good reason leave f(i=-1, i=-1) as UB.
harmic and supercat provide the main answers that provide a reason for the UB. Harmic points out that an optimizing compiler that might break up the ostensibly atomic assignment operations into multiple machine instructions, and that it might further interleave those instructions for optimal speed. This could lead to some very surprising results: i ends up as -2 in his scenario! Thus, harmic demonstrates how assigning the same value to a variable more than once can have ill effects if the operations are unsequenced.
supercat provides a related exposition of the pitfalls of trying to get f(i=-1, i=-1) to do what it looks like it ought to do. He points out that on some architectures, there are hard restrictions against multiple simultaneous writes to the same memory address. A compiler could have a hard time catching this if we were dealing with something less trivial than f(i=-1, i=-1).
davidf also provides an example of interleaving instructions very similar to harmic's.
Although each of harmic's, supercat's and davidf' examples are somewhat contrived, taken together they still serve to provide a tangible reason why f(i=-1, i=-1) should be undefined behavior.
I accepted harmic's answer because it did the best job of addressing all meanings of why, even though Paul Draper's answer addressed the "for what cause" portion better.
other answers
JohnB points out that if we consider overloaded assignment operators (instead of just plain scalars), then we can run into trouble as well.
Since the operations are unsequenced, there is nothing to say that the instructions performing the assignment cannot be interleaved. It might be optimal to do so, depending on CPU architecture. The referenced page states this:
If A is not sequenced before B and B is not sequenced before A, then
two possibilities exist:
evaluations of A and B are unsequenced: they may be performed in any order and may overlap (within a single thread of execution, the
compiler may interleave the CPU instructions that comprise A and B)
evaluations of A and B are indeterminately-sequenced: they may be performed in any order but may not overlap: either A will be complete
before B, or B will be complete before A. The order may be the
opposite the next time the same expression is evaluated.
That by itself doesn't seem like it would cause a problem - assuming that the operation being performed is storing the value -1 into a memory location. But there is also nothing to say that the compiler cannot optimize that into a separate set of instructions that has the same effect, but which could fail if the operation was interleaved with another operation on the same memory location.
For example, imagine that it was more efficient to zero the memory, then decrement it, compared with loading the value -1 in. Then this:
f(i=-1, i=-1)
might become:
clear i
clear i
decr i
decr i
Now i is -2.
It is probably a bogus example, but it is possible.
First, "scalar object" means a type like a int, float, or a pointer (see What is a scalar Object in C++?).
Second, it may seem more obvious that
f(++i, ++i);
would have undefined behavior. But
f(i = -1, i = -1);
is less obvious.
A slightly different example:
int i;
f(i = 1, i = -1);
std::cout << i << "\n";
What assignment happened "last", i = 1, or i = -1? It's not defined in the standard. Really, that means i could be 5 (see harmic's answer for a completely plausible explanation for how this chould be the case). Or you program could segfault. Or reformat your hard drive.
But now you ask: "What about my example? I used the same value (-1) for both assignments. What could possibly be unclear about that?"
You are correct...except in the way the C++ standards committee described this.
If a side effect on a scalar object is unsequenced relative to another side effect on the same scalar object, the behavior is undefined.
They could have made a special exception for your special case, but they didn't. (And why should they? What use would that ever possibly have?) So, i could still be 5. Or your hard drive could be empty. Thus the answer to your question is:
It is undefined behavior because it is not defined what the behavior is.
(This deserves emphasis because many programmers think "undefined" means "random", or "unpredictable". It doesn't; it means not defined by the standard. The behavior could be 100% consistent, and still be undefined.)
Could it have been defined behavior? Yes. Was it defined? No. Hence, it is "undefined".
That said, "undefined" doesn't mean that a compiler will format your hard drive...it means that it could and it would still be a standards-compliant compiler. Realistically, I'm sure g++, Clang, and MSVC will all do what you expected. They just wouldn't "have to".
A different question might be Why did the C++ standards committee choose to make this side-effect unsequenced?. That answer will involve history and opinions of the committee. Or What is good about having this side-effect unsequenced in C++?, which permits any justification, whether or not it was the actual reasoning of the standards committee. You could ask those questions here, or at programmers.stackexchange.com.
A practical reason to not make an exception from the rules just because the two values are the same:
// config.h
#define VALUEA 1
// defaults.h
#define VALUEB 1
// prog.cpp
f(i = VALUEA, i = VALUEB);
Consider the case this was allowed.
Now, some months later, the need arises to change
#define VALUEB 2
Seemingly harmless, isn't it? And yet suddenly prog.cpp wouldn't compile anymore.
Yet, we feel that compilation should not depend on the value of a literal.
Bottom line: there is no exception to the rule because it would make successful compilation depend on the value (rather the type) of a constant.
EDIT
#HeartWare pointed out that constant expressions of the form A DIV B are not allowed in some languages, when B is 0, and cause compilation to fail. Hence changing of a constant could cause compilation errors in some other place. Which is, IMHO, unfortunate. But it is certainly good to restrict such things to the unavoidable.
The confusion is that storing a constant value into a local variable is not one atomic instruction on every architecture the C is designed to be run on. The processor the code runs on matters more than the compiler in this case. For example, on ARM where each instruction can not carry a complete 32 bits constant, storing an int in a variable needs more that one instruction. Example with this pseudo code where you can only store 8 bits at a time and must work in a 32 bits register, i is a int32:
reg = 0xFF; // first instruction
reg |= 0xFF00; // second
reg |= 0xFF0000; // third
reg |= 0xFF000000; // fourth
i = reg; // last
You can imagine that if the compiler wants to optimize it may interleave the same sequence twice, and you don't know what value will get written to i; and let's say that he is not very smart:
reg = 0xFF;
reg |= 0xFF00;
reg |= 0xFF0000;
reg = 0xFF;
reg |= 0xFF000000;
i = reg; // writes 0xFF0000FF == -16776961
reg |= 0xFF00;
reg |= 0xFF0000;
reg |= 0xFF000000;
i = reg; // writes 0xFFFFFFFF == -1
However in my tests gcc is kind enough to recognize that the same value is used twice and generates it once and does nothing weird. I get -1, -1
But my example is still valid as it is important to consider that even a constant may not be as obvious as it seems to be.
Behavior is commonly specified as undefined if there is some conceivable reason why a compiler which was trying to be "helpful" might do something which would cause totally unexpected behavior.
In the case where a variable is written multiple times with nothing to ensure that the writes happen at distinct times, some kinds of hardware might allow multiple "store" operations to be performed simultaneously to different addresses using a dual-port memory. However, some dual-port memories expressly forbid the scenario where two stores hit the same address simultaneously, regardless of whether or not the values written match. If a compiler for such a machine notices two unsequenced attempts to write the same variable, it might either refuse to compile or ensure that the two writes cannot get scheduled simultaneously. But if one or both of the accesses is via a pointer or reference, the compiler might not always be able to tell whether both writes might hit the same storage location. In that case, it might schedule the writes simultaneously, causing a hardware trap on the access attempt.
Of course, the fact that someone might implement a C compiler on such a platform does not suggest that such behavior shouldn't be defined on hardware platforms when using stores of types small enough to be processed atomically. Trying to store two different values in unsequenced fashion could cause weirdness if a compiler isn't aware of it; for example, given:
uint8_t v; // Global
void hey(uint8_t *p)
{
moo(v=5, (*p)=6);
zoo(v);
zoo(v);
}
if the compiler in-lines the call to "moo" and can tell it doesn't modify
"v", it might store a 5 to v, then store a 6 to *p, then pass 5 to "zoo",
and then pass the contents of v to "zoo". If "zoo" doesn't modify "v",
there should be no way the two calls should be passed different values,
but that could easily happen anyway. On the other hand, in cases where
both stores would write the same value, such weirdness could not occur and
there would on most platforms be no sensible reason for an implementation
to do anything weird. Unfortunately, some compiler writers don't need any
excuse for silly behaviors beyond "because the Standard allows it", so even
those cases aren't safe.
C++17 defines stricter evaluation rules. In particular, it sequences function arguments (although in unspecified order).
N5659 §4.6:15
Evaluations A and B are indeterminately sequenced when either A is sequenced before B or B is sequenced before A,
but it is unspecified which. [ Note: Indeterminately sequenced evaluations cannot overlap, but either could
be executed first. —end note ]
N5659 § 8.2.2:5
The
initialization of a parameter, including every associated value computation and side effect, is indeterminately
sequenced with respect to that of any other parameter.
It allows some cases which would be UB before:
f(i = -1, i = -1); // value of i is -1
f(i = -1, i = -2); // value of i is either -1 or -2, but not specified which one
The fact that the result would be the same in most implementations in this case is incidental; the order of evaluation is still undefined. Consider f(i = -1, i = -2): here, order matters. The only reason it doesn't matter in your example is the accident that both values are -1.
Given that the expression is specified as one with an undefined behaviour, a maliciously compliant compiler might display an inappropriate image when you evaluate f(i = -1, i = -1) and abort the execution - and still be considered completely correct. Luckily, no compilers I am aware of do so.
It looks to me like the only rule pertaining to sequencing of function argument expression is here:
3) When calling a function (whether or not the function is inline, and whether or not explicit function call syntax is used), 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.
This does not define sequencing between argument expressions, so we end up in this case:
1) If a side effect on a scalar object is unsequenced relative to another side effect on the same scalar object, the behavior is undefined.
In practice, on most compilers, the example you quoted will run fine (as opposed to "erasing your hard disk" and other theoretical undefined behavior consequences).
It is, however, a liability, as it depends on specific compiler behaviour, even if the two assigned values are the same. Also, obviously, if you tried to assign different values, the results would be "truly" undefined:
void f(int l, int r) {
return l < -1;
}
auto b = f(i = -1, i = -2);
if (b) {
formatDisk();
}
The assignment operator could be overloaded, in which case the order could matter:
struct A {
bool first;
A () : first (false) {
}
const A & operator = (int i) {
first = !first;
return * this;
}
};
void f (A a1, A a2) {
// ...
}
// ...
A i;
f (i = -1, i = -1); // the argument evaluated first has ax.first == true
Actually, there's a reason not to depend on the fact that compiler will check that i is assigned with the same value twice, so that it's possible to replace it with single assignment. What if we have some expressions?
void g(int a, int b, int c, int n) {
int i;
// hey, compiler has to prove Fermat's theorem now!
f(i = 1, i = (ipow(a, n) + ipow(b, n) == ipow(c, n)));
}
This is just answering the "I'm not sure what "scalar object" could mean besides something like an int or a float".
I would interpret the "scalar object" as a abbreviation of "scalar type object", or just "scalar type variable". Then, pointer, enum (constant) are of scalar type.
This is a MSDN article of Scalar Types.
I've (essentially) come across the following in the wild
x = x = 5;
which apparently compiles cleanly under earlier version of gcc (generates a warning under gcc 4.5.1). As far as I can tell the warning is generated by -Wsequence-point. So my question is does this violate the wording in the standard about manipulating variables in between sequence points (i.e., it is undefined behavior per the spec) or is this a gcc false positive (i.e., it is defined behavior per the spec)? The wording on sequence points is a bit hard to follow.
I said essentially because what I actually came across (in a larger expression) was
x[0][0] = x[0][0] = 5;
but I didn't think that was material to the warning (please correct me if that is to the point and not what I've assumed is the crux of the matter).
Assuming x is of built-in type, it assigns to x twice without an intervening sequence point, which is all you need to know. The fact that both assignments are of the same value (5), and could in theory be optimized into a single assignment (if x is not volatile), is neither here nor there.
At least, that's how I interpret "modified" in the standard - assigned a value, regardless of whether it happens to be the same as the old value. Likewise, casting away const and assigning to a const object is, I think, UB regardless of whether the value you assign happens to be equal to the prior value. Otherwise there's be a huge overhead on all memory writes if an implementation wanted to put string literals into ROM, to prevent a page fault in that case, and we know by inspection that compilers don't emit that code.
A even more exciting example would be x[0][0] = x[0][i] = 5;, which assigns to the same object without an intervening sequence point if (and only if) i == 0, so is defined behaviour conditional on the value of i.
I don't see quite why a compiler might do anything unexpected in either case, but again my lack of imagination is irrelevant :-)
What ablenky says is right. If you're in some context where you can't use two statements, maybe write x[0][0] = 5, x[0][i] = 5 instead. In both your given cases, just ditch the redundant assignment.
Consider the following C++ Standard ISO/IEC 14882:2003(E) citation (section 5, paragraph 4):
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. 53) 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 unspecified
i = 7, i++, i++; // i becomes 9
i = ++i + 1; // the behavior is unspecified
i = i + 1; // the value of i is incremented
—end example]
I was surprised that i = ++i + 1 gives an undefined value of i.
Does anybody know of a compiler implementation which does not give 2 for the following case?
int i = 0;
i = ++i + 1;
std::cout << i << std::endl;
The thing is that operator= has two args. First one is always i reference.
The order of evaluation does not matter in this case.
I do not see any problem except C++ Standard taboo.
Please, do not consider such cases where the order of arguments is important to evaluation. For example, ++i + i is obviously undefined. Please, consider only my case
i = ++i + 1.
Why does the C++ Standard prohibit such expressions?
You make the mistake of thinking of operator= as a two-argument function, where the side effects of the arguments must be completely evaluated before the function begins. If that were the case, then the expression i = ++i + 1 would have multiple sequence points, and ++i would be fully evaluated before the assignment began. That's not the case, though. What's being evaluated in the intrinsic assignment operator, not a user-defined operator. There's only one sequence point in that expression.
The result of ++i is evaluated before the assignment (and before the addition operator), but the side effect is not necessarily applied right away. The result of ++i + 1 is always the same as i + 2, so that's the value that gets assigned to i as part of the assignment operator. The result of ++i is always i + 1, so that's what gets assigned to i as part of the increment operator. There is no sequence point to control which value should get assigned first.
Since the code is violating the rule that "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," the behavior is undefined. Practically, though, it's likely that either i + 1 or i + 2 will be assigned first, then the other value will be assigned, and finally the program will continue running as usual — no nasal demons or exploding toilets, and no i + 3, either.
It's undefined behaviour, not (just) unspecified behaviour because there are two writes to i without an intervening sequence point. It is this way by definition as far as the standard specifies.
The standard allows compilers to generate code that delays writes back to storage - or from another view point, to resequence the instructions implementing side effects - any way it chooses so long as it complies with the requirements of sequence points.
The issue with this statement expression is that it implies two writes to i without an intervening sequence point:
i = i++ + 1;
One write is for the value of the original value of i "plus one" and the other is for that value "plus one" again. These writes could happen in any order or blow up completely as far as the standard allows. Theoretically this even gives implementations the freedom to perform writebacks in parallel without bothering to check for simultaneous access errors.
C/C++ defines a concept called sequence points, which refer to a point in execution where it's guaranteed that all effects of previous evaluations will have already been performed. Saying i = ++i + 1 is undefined because it increments i and also assigns i to itself, neither of which is a defined sequence point alone. Therefore, it is unspecified which will happen first.
Update for C++11 (09/30/2011)
Stop, this is well defined in C++11. It was undefined only in C++03, but C++11 is more flexible.
int i = 0;
i = ++i + 1;
After that line, i will be 2. The reason for this change was ... because it already works in practice and it would have been more work to make it be undefined than to just leave it defined in the rules of C++11 (actually, that this works now is more of an accident than a deliberate change, so please don't do it in your code!).
Straight from the horse's mouth
http://www.open-std.org/jtc1/sc22/wg21/docs/cwg_defects.html#637
Given two choices: defined or undefined, which choice would you have made?
The authors of the standard had two choices: define the behavior or specify it as undefined.
Given the clearly unwise nature of writing such code in the first place, it doesn't make any sense to specify a result for it. One would want to discourage code like that and not encourage it. It's not useful or necessary for anything.
Furthermore, standards committees do not have any way to force compiler writers to do anything. Had they required a specific behavior it is likely that the requirement would have been ignored.
There are practical reasons as well, but I suspect they were subordinate to the above general consideration. But for the record, any sort of required behavior for this kind of expression and related kinds will restrict the compiler's ability to generate code, to factor out common subexpressions, to move objects between registers and memory, etc. C was already handicapped by weak visibility restrictions. Languages like Fortran long ago realized that aliased parameters and globals were an optimization-killer and I believe they simply prohibited them.
I know you were interested in a specific expression, but the exact nature of any given construct doesn't matter very much. It's not going to be easy to predict what a complex code generator will do and the language attempts to not require those predictions in silly cases.
The important part of the standard is:
its stored value modified at most once by the evaluation of an expression
You modify the value twice, once with the ++ operator, once with the assignment
Please note that your copy of the standard is outdated and contains a known (and fixed) error just in 1st and 3rd code lines of your example, see:
C++ Standard Core Language Issue Table of Contents, Revision 67, #351
and
Andrew Koenig: Sequence point error: unspecified or undefined?
The topic is not easy to get just reading the standard (which is pretty obscure :( in this case).
For example, will it be well(or not)-defined, unspecified or else in general case actually depends not only on the statement structure, but also on memory contents (to be specific, variable values) at the moment of execution, another example:
++i, ++i; //ok
(++i, ++j) + (++i, ++j); //ub, see the first reference below (12.1 - 12.3)
Please have a look at (it has it all clear and precise):
JTC1/SC22/WG14 N926 "Sequence Point Analysis"
Also, Angelika Langer has an article on the topic (though not as clear as the previous one):
"Sequence Points and Expression Evaluation in C++"
There was also a discussion in Russian (though with some apparently erroneous statements in the comments and in the post itself):
"Точки следования (sequence points)"
The following code demonstrates how you could get the wrong(unexpected) result:
int main()
{
int i = 0;
__asm { // here standard conformant implementation of i = ++i + 1
mov eax, i;
inc eax;
mov ecx, 1;
add ecx, eax;
mov i, ecx;
mov i, eax; // delayed write
};
cout << i << endl;
}
It will print 1 as a result.
Assuming you are asking "Why is the language designed this way?".
You say that i = ++i + i is "obviously undefined" but i = ++i + 1 should leave i with a defined value? Frankly, that would not be very consistent. I prefer to have either everything perfectly defined, or everything consistently unspecified. In C++ I have the latter. It's not a terribly bad choice per se - for one thing, it prevents you from writing evil code which makes five or six modifications in the same "statement".
Argument by analogy: If you think of operators as types of functions, then it kind of makes sense. If you had a class with an overloaded operator=, your assignment statement would be equivalent to something like this:
operator=(i, ++i+1)
(The first parameter is actually passed in implicitly via the this pointer, but this is just for illustration.)
For a plain function call, this is obviously undefined. The value of the first argument depends on when the second argument is evaluated. However with primitive types you get away with it because the original value of i is simply overwritten; its value doesn't matter. But if you were doing some other magic in your own operator=, then the difference could surface.
Simply put: all operators act like functions, and should therefore behave according to the same notions. If i + ++i is undefined, then i = ++i should be undefined as well.
How about, we just all agree to never, never, write code like this? If the compiler doesn't know what you want to do, how do you expect the poor sap that is following on behind you to understand what you wanted to do? Putting i++; on it's own line will not kill you.
The underlying reason is because of the way the compiler handles reading and writing of values. The compiler is allowed to store an intermediate value in memory and only actually commit the value at the end of the expression. We read the expression ++i as "increase i by one and return it", but a compiler might see it as "load the value of i, add one, return it, and the commit it back to memory before someone uses it again. The compiler is encouraged to avoid reading/writing to the actual memory location as much as possible, because that would slow the program down.
In the specific case of i = ++i + 1, it suffers largely due to the need of consistent behavioral rules. Many compilers will do the 'right thing' in such a situation, but what if one of the is was actually a pointer, pointing to i? Without this rule, the compiler would have to be very careful to make sure it performed the loads and stores in the right order. This rule serves to allow for more optimization opportunities.
A similar case is that of the so-called strict-aliasing rule. You can't assign a value (say, an int) through a value of an unrelated type (say, a float) with only a few exceptions. This keeps the compiler from having to worry that some float * being used will change the value of an int, and greatly improves optimization potential.
The problem here is that the standard allows a compiler to completely reorder a statement while it is executing. It is not, however, allowed to reorder statements (so long as any such reordering results in changed program behavior). Therefore, the expression i = ++i + 1; may be evaluated two ways:
++i; // i = 2
i = i + 1;
or
i = i + 1; // i = 2
++i;
or
i = i + 1; ++i; //(Running in parallel using, say, an SSE instruction) i = 1
This gets even worse when you have user defined types thrown in the mix, where the ++ operator can have whatever effect on the type the author of the type wants, in which case the order used in evaluation matters significantly.
i = v[i++]; // the behavior is unspecified
i = ++i + 1; // the behavior is unspecified
All the above expressions invoke Undefined Behavior.
i = 7, i++, i++; // i becomes 9
This is fine.
Read Steve Summit's C-FAQs.
From ++i, i must assigned "1", but with i = ++i + 1, it must be assigned the value "2". Since there is no intervening sequence point, the compiler can assume that the same variable is not being written twice, so this two operations can be done in any order. so yes, the compiler would be correct if the final value is 1.