Consider a class Foo of which foo is an instance.
Foo has a function calculate() that changes an internal member for which watermark() is the "getter", and returns something with the same type as that internal member.
Is the expression
foo.calculate() != foo.watermark()
well-defined. That is, must calculate() happen before watermark() is called?
In order words, is the evaluation order strictly foo.calculate(), followed by foo.watermark() followed by !=?
There are no sequential point between both, the order of evaluation is unspecified.
You have to force sequence as for example:
const auto& res = foo.calculate();
res != foo.watermark();
The C++ Standard(ISO/IEC 14882:2014) say's :
Sequenced before is an asymmetric, transitive, pair-wise relation
between evaluations executed by a single thread, which induces a
partial order among those evaluations. Given any two evaluations A and
B, if A is sequenced before B, then the execution of A shall precede
the execution of B. If A is not sequenced before B and B is not
sequenced before A, then A and B are unsequenced. 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.
Do not allow the same scalar object to appear in side effects or value computations in both halves of an unsequenced or indeterminately sequenced operation.
Reference link here.
It may help: http://en.cppreference.com/w/cpp/language/eval_order
Order of evaluation of the operands of almost all C++ operators (including the order of evaluation of function arguments in a function-call expression and the order of evaluation of the subexpressions within any expression) is unspecified. The compiler can evaluate operands in any order, and may choose another order when the same expression is evaluated again.
I've looked at a bunch of questions regarding sequence points, and haven't been able to figure out if the order of evaluation for x*f(x) is guaranteed if f modifies x, and is this different for f(x)*x.
Consider this code:
#include <iostream>
int fx(int &x) {
x = x + 1;
return x;
}
int f1(int &x) {
return fx(x)*x; // Line A
}
int f2(int &x) {
return x*fx(x); // Line B
}
int main(void) {
int a = 6, b = 6;
std::cout << f1(a) << " " << f2(b) << std::endl;
}
This prints 49 42 on g++ 4.8.4 (Ubuntu 14.04).
I'm wondering whether this is guaranteed behavior or unspecified.
Specifically, in this program, fx gets called twice, with x=6 both times, and returns 7 both times. The difference is that Line A computes 7*7 (taking the value of x after fx returns) while Line B computes 6*7 (taking the value of x before fx returns).
Is this guaranteed behavior? If yes, what part of the standard specifies this?
Also: If I change all the functions to use int *x instead of int &x and make corresponding changes to places they're called from, I get C code which has the same issues. Is the answer any different for C?
In terms of evaluation sequence, it is easier to think of x*f(x) as if it was:
operator*(x, f(x));
so that there are no mathematical preconceptions on how multiplication is supposed to work.
As #dan04 helpfully pointed out, the standard says:
Section 1.9.15: “Except where noted, evaluations of operands of individual operators and of subexpressions of individual expressions are unsequenced.”
This means that the compiler is free to evaluate these arguments in any order, the sequence point being operator* call. The only guarantee is that before the operator* is called, both arguments have to be evaluated.
In your example, conceptually, you could be certain that at least one of the arguments will be 7, but you cannot be certain that both of them will. To me, this would be enough to label this behaviour as undefined; however, #user2079303 answer explains well why it is not technically the case.
Regardless of whether the behaviour is undefined or indeterminate, you cannot use such an expression in a well-behaved program.
The evaluation order of arguments is not specified by the standard, so the behaviour that you see is not guaranteed.
Since you mention sequence points, I'll consider the c++03 standard which uses that term while the later standards have changed wording and abandoned the term.
ISO/IEC 14882:2003(E) §5 /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...
There is also discussion on whether this is undefined behaviour or is the order merely unspecified. The rest of that paragraph sheds some light (or doubt) on that.
ISO/IEC 14882:2003(E) §5 /4:
... 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.
x is indeed modified in f and it's value is read as an operand in the same expression where f is called. And it's not specified whether x reads the modified or non-modified value. That might scream Undefined Behaviour! to you, but hold your horses, because the standard also states:
ISO/IEC 14882:2003(E) §1.9 /17:
... When calling a function (whether or not the function is inline), there is a sequence point after the evaluation of all function arguments (if any) which takes place before execution of any expressions or statements in the function body. There is also a sequence point after the copying of a returned value and before the execution of any expressions outside the function 11) ...
So, if f(x) is evaluated first, then there is a sequence point after copying the returned value. So the above rule about UB does not apply because the read of x is not between the next and previous sequence point. The x operand will have the modified value.
If x is evaluated first, then there is a sequence point after evaluating the arguments of f(x) Again, the rule about UB does not apply. In this case x operand will have the non-modified value.
In summary, the order is unspecified but there is no undefined behaviour. It's a bug, but the outcome is predictable to some degree. The behaviour is the same in the later standards, even though the wording changed. I'll not delve into those since it's already covered well in other good answers.
Since you ask about similar situation in C
C89 (draft) 3.3/3:
Except as indicated by the syntax 27 or otherwise specified later (for the function-call operator () , && , || , ?: , and comma operators), the order of evaluation of subexpressions and the order in which side effects take place are both unspecified.
The function call exception is already mentioned here. Following is the paragraph that implies the undefined behaviour if there were no sequence points:
C89 (draft) 3.3/2:
Between the previous and next sequence point an 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.26
And here are the sequence points defined:
C89 (draft) A.2
The following are the sequence points described in 2.1.2.3
The call to a function, after the arguments have been evaluated (3.3.2.2).
...
... the expression in a return statement (3.6.6.4).
The conclusions are the same as in C++.
A quick note on something I don't see covered explicitly by the other answers:
if the order of evaluation for x*f(x) is guaranteed if f modifies x, and is this different for f(x)*x.
Consider, as in Maksim's answer
operator*(x, f(x));
now there are only two ways of evaluating both arguments before the call as required:
auto lhs = x; // or auto rhs = f(x);
auto rhs = f(x); // or auto lhs = x;
return lhs * rhs
So, when you ask
I'm wondering whether this is guaranteed behavior or unspecified.
the standard doesn't specify which of those two behaviours the compiler must choose, but it does specify those are the only valid behaviours.
So, it's neither guaranteed nor entirely unspecified.
Oh, and:
I've looked at a bunch of questions regarding sequence points, and haven't been able to figure out if the order of evaluation ...
sequence points are a used in the C language standard's treatment of this, but not in the C++ standard.
In the expression x * y, the terms x and y are unsequenced. This is one of the three possible sequencing relations, which are:
A sequenced-before B: A must be evaluated, with all side-effects complete, before B begins evaluationg
A and B indeterminately-sequenced: one of the two following cases is true: A is sequenced-before B, or B is sequenced-before A. It is unspecified which of those two cases holds.
A and B unsequenced: There is no sequencing relation defined between A and B.
It is important to note that these are pair-wise relations. We cannot say "x is unsequenced". We can only say that two operations are unsequenced with respect to each other.
Also important is that these relations are transitive; and the latter two relations are symmetric.
unspecified is a technical term which means that the Standard specifies a set number of possible results. This is different to undefined behaviour which means that the Standard does not cover the behaviour at all. See here for further reading.
Moving onto the code x * f(x). This is identical to f(x) * x, because as discussed above, x and f(x) are unsequenced, with respect to each other, in both cases.
Now we come to the point where several people seem to be coming unstuck. Evaluating the expression f(x) is unsequenced with respect to x. However, it does not follow that any statements inside the function body of f are also unsequenced with respect to x. In fact, there are sequencing relations surrounding any function call, and those relations cannot be ignored.
Here is the text from C++14:
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. [Note: Value computations and side effects associated with different argument expressions are unsequenced. —end note ]
Every evaluation in the calling function (including other function calls) that is not otherwise specifically sequenced before or after the execution of the body of the called function is indeterminately sequenced with
respect to the execution of the called function.
with footnote:
In other words, function executions do not interleave with each other.
The bolded text clearly states that for the two expressions:
A: x = x + 1; inside f(x)
B: evaluating the first x in the expression x * f(x)
their relationship is: indeterminately sequenced.
The text regarding undefined behaviour and sequencing is:
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.
In this case, the relation is indeterminately sequenced, not unsequenced. So there is no undefined behaviour.
The result is instead unspecified according to whether x is sequenced before x = x + 1 or the other way around. So there are only two possible outcomes, 42 and 49.
In case anyone had qualms about the x in f(x), the following text applies:
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.
So the evaluation of that x is sequenced before x = x + 1. This is an example of an evlauation that falls under the case of "specifically sequenced before" in the bolded quote above.
Footnote: the behaviour was exactly the same in C++03, but the terminology was different. In C++03 we say that there is a sequence point upon entry and exit of every function call, therefore the write to x inside the function is separated from the read of x outside the function by at least one sequence point.
You need to distinguish:
a) Operator precedence and associativity, which controls the order in which the values of subexpressions are combined by their operators.
b) The sequence of subexpression evaluation. E.g. in the expression f(x)/g(x), the compiler can evaluate g(x) first and f(x) afterwards. Nonetheless, the resulting value must be computed by dividing respective sub-values in the right order, of course.
c) The sequence of side-effects of the subexpressions. Roughly speaking, for example, the compiler might, for sake of optimization, decide to write values to the affected variables only at the end of the expression or any other suitable place.
As a very rough approximation, you can say, that within a single expression, the order of evaluation (not associativity etc.) is more or less unspecified. If you need a specific order of evaluation, break down the expression into series of statements like this:
int a = f(x);
int b = g(x);
return a/b;
instead of
return f(x)/g(x);
For exact rules, see http://en.cppreference.com/w/cpp/language/eval_order
Order of evaluation of the operands of almost all C++ operators is
unspecified. The compiler can evaluate operands in any order, and may
choose another order when the same expression is evaluated again
As the order of evaluation is not always the same hence you may get unexpected results.
Order of evaluation
For example:
int foo(int i) { return i; }
int main()
{
int i = 0;
i = i++; // Undefined
i = foo(i++); // ?
return 0;
}
What would the current ISO C++ standard specify for this case?
EDIT:
Here's where I get confused:
Except where noted, evaluations of operands of individual operators and of subexpressions of individual expressions are unsequenced.
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.
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
Every evaluation in the calling function (including other function calls) that is not otherwise specifically
sequenced before or after the execution of the body of the called function is indeterminately sequenced with
respect to the execution of the called function.
So it seems you could have a value computation on the left side of the assignment (just i), and a side effect on the right side (the modification of i from i++) which aren't sequenced with respect to each other.
EDIT2:
For anyone who finds themselves here, there is a really great explanation about sequencing that I found here.
The last sentence in your quote says "that is not otherwise specifically sequenced before or after the execution of the body of the called function" so the question is whether the increment and the assignment are "otherwise specifically sequenced before or after" the function body.
1.9 [intro.execution] p15 has the answer:
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. [ Note: Value
computations and side effects associated with different argument expressions are unsequenced. — end note ]
So the increment of i happens before the function body, and the assignment to i happens after the function returns, so it is perfectly well-defined.
In pre-C++11 terminology, the function call introduces a sequence point between the increment and the assignment.
i = foo(i++); is fine, because i++ is executed before foo() is called. A copy of i is made, i is then incremented, then the copy is passed to foo(). It is the same as doing this explicitly:
int tmp = i++;
i = foo(tmp);
This question already has answers here:
What is the correct answer for cout << a++ << a;?
(4 answers)
Closed 7 years ago.
#include <iostream>
using namespace std;
int main()
{
int x=80;
int &y=x;
cout<<"x"<<x<<" "<<"y"<<y++;
return 0;
}
The above code gave me the following output:
81 80
Can anyone explain me how the value of x changes to 81? The value of y is 80 and it later gets incremented to 81, but how did it reflect in x?
Did it reflect because y is a reference variable? Then the value should have been modified in both x and y?
You have undefined behaviour because your operations are between two consecutive sequence points (there are no sequence points between function arguments evaluation). You can loosely think of sequence points as "temporal" markers and between two consecutive ones you are not allowed to modify the same variable more than once.
Basically your code is equivalent to
std::cout << x << x++; // undefined behaviour
since y is just a reference (an alias) for x.
1.9 Program execution [intro.execution] (emphasize mine)
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) 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, and they are not potentially concurrent (1.10),
the behavior is undefined. [ Note: The next section imposes similar,
but more complex restrictions on potentially concurrent computations.
—endnote]
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. [ Note: Value
computations and side effects associated with different argument
expressions are unsequenced. — end note ] Every evaluation in the
calling function (including other function calls) that is not
otherwise specifically sequenced before or after the execution of the
body of the called function is indeterminately sequenced with respect
to the execution of the called function.9 Several contexts in C++
cause evaluation of a function call, even though no corresponding
function call syntax appears in the translation unit. [ Example:
Evaluation of a new-expression invokes one or more allocation and
constructor functions; see 5.3.4. For another example, invocation of a
conversion function (12.3.2) can arise in contexts in which no
function call syntax appears.
Related: https://stackoverflow.com/a/10782972/3093378
Suppose I have types T and U, and functions U f(T) and T g(), and I write the expression f(g()). Under what circumstances is it possible that code in nearby unsequenced expressions may execute after g but before f?
I understand that one circumstance is in a function call like h(f(g()), j()), j may execute at any time relative to g and f. Is this essentially the only example, or are there others?
For motivation, the functions std::make_shared<T> and std::make_unique<T> can be used to write more exception-safe code, as shown in this example of unsafe code from http://herbsutter.com/gotw/_102/:
// In some header file:
void f( std::unique_ptr<T1>, std::unique_ptr<T2> );
// At some call site:
f( std::unique_ptr<T1>{ new T1 }, std::unique_ptr<T2>{ new T2 } );
The T2 constructor may throw an exception after the T1 constructor but before the std::unique_ptr<T1> constructor, causing the T1 to be leaked. The solution is to write f( make_unique<T1>(), make_unique<T2>() ); instead.
Every discussion I've found of the exception safety provided by std::unique_ptr uses the same example. This makes me wonder if argument expressions in multiple-argument functions (including certain operators like + and []) are the only situation in which this behavior is expected.
[intro.execution] (§1.9) contains the rules, which are actually pretty simple.
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 Except where noted, evaluations of operands of individual operators and of subexpressions of individual expressions are unsequenced.
A full expression is just what it says it is: an expression which is not a subexpression of any other expression. So only subexpressions of an expression can be unsequenced, and not all of them. For example, certain operators sequence the execution of their arguments, and function calls are indeterminately sequenced (that is, not interleaved).