Isn't the evaluated value different from the expression? i++ returns the value of i before increasing. i-=-1 would return the value after it has been increased. Wouldn't it be more correct to make it equal to ++i

[–] Jamie@jamie.moe 2 points 1 year ago (1 children)

I gave it a shot in Compiler Explorer, with the following code:

#include <stdio.h>

int main() {
  for (int i = 0; i < 10; i -= -1) {
    printf("%d", i);
  }
}

GCC takes the i-=-1 and optimizes it into ADD DWARD PTR [rbp-4], 1, and changing it around to ++i or i++ makes no difference.

So, at least in C and C++, it works all the same. Even on unsigned integers.

[–] mfz@kbin.social 3 points 1 year ago

It works the same because the value of the last expression in the for loop is not used for anything. It's the side effect of that statement that counts. Eg, the value of i is checked the next time the for loop is executed by the condition check. Try replacing i in the condition check instead with i++ or ++i and you would see different results.

Something like: for (int i = 0; ++i < 10;) { ... }

load more comments (9 replies)