4 ms·
It's to have the compiler produce a warning if the types aren't the same. __cmp_once produces such a warning, and they don't want the warning to go away if it
by LukeShu 9y ago
It's to have the compiler produce a warning if the types aren't the same. __cmp_once produces such a warning, and they don't want the warning to go away if it decides to use plain __cmp instead of __cmp_once.
It doesn't "do" anything otherwise; it always evaluates to "1".
- gshrikant 9y agoDoesn't the compiler produce a warning by default (on comparing a `char` and `int`, for example) when using -Wall?