Different output for same code in C++14 and c++17

Question

I am getting different answers for the same c++ code but in different versions of c++ i.e. c++14 and c++17

What are the changed made into c++17 from c++14 because of which I am getting different answers ?

(Specifically related to this question)

#include<iostream>

using namespace std;

int main()
{
    int i = 1;
    cout << i++ <<" "<< i-- << " " << i--;
    return 0;
}

/*
* Output in c++17
*   1 2 1
*
* Output in c++14
*  -1 0 1
*
*/

@DYZ Yes, it's UB - multiple unsequenced modifications to the variable, `i`. — Adrian Mole, Apr 01 '20 at 19:51
It is. Changing a variable while using it in the same command is undefined behavior and the result can't be predicted. The change might be because of compiler core changes. — Roy Avidan, Apr 01 '20 at 19:51
fwiw, even if this was well defined you shouldn't be writing code like that — 463035818_is_not_a_number, Apr 01 '20 at 19:53
But wasn't order of evaluation fixed with C++17? Or does it not cover operators? — Yksisarvinen, Apr 01 '20 at 19:53
@Yksisarvinen It was fixed for this operator. Just closed as a dupe to the canonical. — NathanOliver, Apr 01 '20 at 19:54
Not an *exact* duplicate, but very related: [Undefined behavior and sequence points](https://stackoverflow.com/q/4176328/10871073). — Adrian Mole, Apr 01 '20 at 19:55

score 1 · Answer 1 · answered Apr 01 '20 at 19:53

C++17 changed the rules for the order of evaluation of some expression.

Let's rewrite the expression to its equivalent in function calls:

std::operator<<(std::operator<<(std::cout.operator<<(i++), " ").operator<<(i--), " ").operator<<(i--);

In C++14, the compiler was permitted to evaluate from right to left, or in ony nested order

To support chaining properly, C++17 states that such expressions are sequenced.

For example, std::operator<<(std::cout.operator<<(i++), " ").operator<<(i--) C++17 requires that i++ is evaluated before i--.

Different output for same code in C++14 and c++17

1 Answers1