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

[–] [email protected] 12 points 1 year ago (1 children)

And that's why post- and pre-increment is non-existant in Python and Rust. It's an easy source for bugs for a noncritical abbreviation🤷

[–] [email protected] 2 points 1 year ago (1 children)

They're especially also a source of bugs, because they encourage manually incrementing indices and manually accessing array positions, which is almost never actually sensible.

[–] [email protected] 3 points 1 year ago

I love iterators so much.

[–] [email protected] 7 points 1 year ago* (last edited 1 year ago)

If you're hell bend on achieving the goodness of i++ equivalent you could wrap it up like this:
(i-=-1,i-1)

We're talking C here of course.

[–] [email protected] 6 points 1 year ago (1 children)

In the languages I know, i-=-1 or x=3 are not expressions, but rather statements, so they do not evaluate to a value.

So, this would be a compiler error:

a = (x=3)

[–] [email protected] 9 points 1 year ago (1 children)

Well, not all languages allow for fun programming :)

[–] [email protected] 5 points 1 year ago

Sounds like the opposite of fun to me, to have those as expressions...

[–] [email protected] 5 points 1 year ago (1 children)

Please explain in less detail to help me understand, internet friend.

[–] [email protected] 5 points 1 year ago

In C you can group expressions within ( and ) separated with ,. Expressions are evaluated in order and the last expression in the group is the returned value of the group.

[–] [email protected] 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.

[–] [email protected] 4 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;) { ... }