Why is memcmp(a, b, 4) only sometimes optimized to a uint32 comparison?

Question

Given this code:

#include 

int equal4(const char* a, const char* b)
{
    return memcmp(a, b, 4) == 0;
}

int less4(const char* a, const char* b         


        

Answer 1

Endianness is one problem, but signed char is another. For example, consider that the four bytes you compare are 0x207f2020 and 0x20802020. The 80 as signed char is -128, the 7f as signed char is +127. But if you compare the four bytes, no comparison will give you the right order.

Of course you can do an xor with 0x80808080 and then you can just use an unsigned compare.