57
i++
(programming.dev)
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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
Please explain in less detail to help me understand, internet friend.
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
orx=3
are not expressions, but rather statements, so they do not evaluate to a value.So, this would be a compiler error:
Well, not all languages allow for fun programming :)
I gave it a shot in Compiler Explorer, with the following code:
GCC takes the
i-=-1
and optimizes it intoADD DWARD PTR [rbp-4], 1
, and changing it around to++i
ori++
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 ofi
is checked the next time the for loop is executed by the condition check. Try replacingi
in the condition check instead withi++
or++i
and you would see different results.Something like:
for (int i = 0; ++i < 10;) { ... }