g++ vs intel/clang argument passing order?

Question

Consider the following code (LWS):

#include <iostream>
#include <chrono>

inline void test(
   const std::chrono::high_resolution_clock::time_point& first, 
   const std::chrono::high_resolution_clock::time_point& second)
{
   std::cout << first.time_since_epoch().count() << std::endl;
   std::cout << second.time_since_epoch().count() << std::endl;
}

int main(int argc, char* argv[])
{
   test(std::chrono::high_resolution_clock::now(), 
        std::chrono::high_resolution_clock::now());
   return 0;
}

You have to run it several times because sometimes, there is no visible difference. But when there is a visible difference between the time of evaluation of first and second, the result is the following under g++ :

1363376239363175
1363376239363174

and the following under intel and clang :

1363376267971435
1363376267971436

It means that under g++, the second argument is evaluated first, and under intel and clang the first argument is evaluated first.

Which one is true according to the C++11 standard?

Answer 1

Which one is true according to the C++11 standard ?

Both are permissible. To quote the standard (§8.3.6):

The order of evaluation of function arguments is unspecified.

Answer 2

I have a slightly simpler example to illustrate the same problem.

bash$ cat test.cpp
#include <iostream>
using namespace std;
int x = 0;
int foo() 
{
    cout << "foo" << endl;
    return x++;
}
int bar()
{
    cout << "bar" << endl;
    return x++;
}
void test_it(int a, int b)
{
    cout << "a = " << a << endl
        << "b = " << b << endl;

}
int main(int argc, const char *argv[])
{
    test_it(foo(),bar()); 
    return 0;
}

bash$ clang++ test.cpp && ./a.out
foo
bar
a = 0
b = 1
bash$ g++ test.cpp && ./a.out
bar
foo
a = 1
b = 0