Effective implementation of conversion small string to uint64_t - c++11

Question

#include <cstdint>
#include <cstring>
template<typename T>
T oph_(const char *s){
constexpr std::size_t MAX = sizeof(T);
const std::size_t size = strnlen(s, MAX);
T r = 0;
for(auto it = s; it - s < size; ++it)
r = r << 8 | *it;
return r;
}
inline uint64_t oph(const char *s){
return oph_<uint64_t>(s);
}
int main(){
uint64_t const a = oph("New York City");
uint64_t const b = oph("Boston International");
return a > b;
}
I want to convert first 8 characters from const char * to uint64_t so I can easily compare if two strings are greater / lesser.
I am aware that equals will semi-work.
However I am not sure if this is most efficient implementation.
I want the implementation to work on both little and big endian machines.

Answer 1

This is a C implementation, that should be faster that your implementation, but I still need to use strncpy which should be the bottleneck
#include <string.h>
#include <stdio.h>
#include <stdint.h>
#include <byteswap.h>
union small_str {
uint64_t v;
char buf[8];
};
static uint64_t fill_small_str(const char *str)
{
union small_str ss = { 0 };
strncpy(ss.buf, str, 8);
#if defined(__BYTE_ORDER__) && (__BYTE_ORDER__ == __ORDER_BIG_ENDIAN__)
return ss.v;
#else
return bswap_64(ss.v);
#endif
}
int main(void)
{
uint64_t const a = fill_small_str("Aew York City");
uint64_t const b = fill_small_str("Boston International");
printf("%lu ; %lu ; %d\n", a, b, (a < b));
return 0;
}