c - are signed integer and unsigned integer one-one mapping with each other? -

i want make ip-port pair search key, have following function

    int64_t make_pair(u_int32_t ip, u_int16_t port)     {             u_int64_t ip_u64 = ip;             ip_u64 = ip_u64 << 16;             int64_t ip_port_pair = (int64_t)(ip_u64 + (u_int64_t)port);             return ip_port_pair;     } 

indeed, want transform u_int64_t int64_t because not convenient unsigned integer compare value. i'm afraid cast u_int64_t int64_t not one-one mapping , there conflicts or search error.

so want ask cast u_int64_t int64_t not one-one mapping or not? thanks!

the c standard guarantees int64_t uses two's-complement representation no padding bits, it's implicitly guaranteed there's one-to-one mapping. (i believe it's still possible negative value trap representation, that's unlikely in practice. unlike ordinary signed integer types, intn_t types cannot make negative value trap representation; 2ⁿ possible representations well-defined , have distinct values.)

on other hand, there's no guarantee int64_t exists.

a conversion uint64_t int64_t yields implementation-defined result values outside range of int64_t. result 1 you'd expect, produced reinterpreting representation, it's not guaranteed.

so in practice, can away converting between int64_t , uint64_t.

but why need to? it's "because not convenient unsigned integer compare value". what's inconvenient it? relational operators defined both signed , unsigned types.