Why does memset take an int instead of a char?

Question

Why does memset take an int as the second argument instead of a char, whereas wmemset takes a wchar_t instea

Answer 1

See fred's answer, it's for performance reasons.

On my side, I tried this code:

#include 
#include 

int main (int argc, const char * argv[])
{
    char c = 0x00;

    printf("Before: c = 0x%02x\n", c);
    memset( &c, 0xABCDEF54, 1);
    printf("After:  c = 0x%02x\n", c);

    return 0;
}

And it gives me this on a 64bits Mac:

Before: c = 0x00
After:  c = 0x54

So as you see, only the last byte gets written. I guess this is dependent on the architecture (endianness).