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
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🤷
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.
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)
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.
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;) { ... }
