Expression evaluation problems in C++

Question

See the following code:

#include<bits/stdc++.h>
using namespace std;
int read(){
    int a;
    scanf("%d", &a);
    return a;
}
void print(int a, int b){
    printf("%d %d", a, b);
}
int main(){
    print(read(), read());
}

The result of the output transitions the order of the numbers, it isn't that C++ call function operation ',' behind?

Please take [the tour](http://stackoverflow.com/tour) and read the [help page](http://stackoverflow.com/help). Here is a nice list of [C++ books](http://stackoverflow.com/questions/388242/the-definitive-c-book-guide-and-list). — , Jan 13 '17 at 14:22
Please read [Why should I not #include ?](http://stackoverflow.com/questions/31816095/why-should-i-not-include-bits-stdc-h). — Some programmer dude, Jan 13 '17 at 14:24
Also, what part (besides that header file) is really "C++"? That looks more like pure C. — Some programmer dude, Jan 13 '17 at 14:24
@Someprogrammerdude it is also valid C++. C++ is a multiparadigm language, one of those is C-style programming — M.M, Jan 13 '17 at 14:26

score 0 · Accepted Answer · answered Jan 13 '17 at 14:24

In general, in C++, the order of evaluation of subexpressions is unspecified. Check out the link for more discussion, and some exceptional cases.

The C++ standard says:

Except where noted, evaluations of operands of individual operators and of subexpressions of individual expressions are unsequenced.[...]

The compiler may decide which argument of print (in your example) to evaluate first, based on efficiency or some other considerations. You should never rely on the order.

Expression evaluation problems in C++

1 Answers1