3 ms·
So what if the code doesn't get called? That doesn't prevent it from being built. Considering that memset() is part of libc, which is a shared library, presum
by humbledrone 16y ago
So what if the code doesn't get called? That doesn't prevent it from being built. Considering that memset() is part of libc, which is a shared library, presumably it's going to get compiled regardless of whether it's called.