在C ++中，这是否可以正确地将两个无符号的32位整数组合为一个无符号的64位整数？

Question

Does this correctly combine two unsigned 32-bit integers into one unsigned 64-bit integer in C++?

std::uint32_t a = ...
std::uint32_t b = ...
std::uint64_t result = ((std::uint64_t)a << 32) | (std::uint64_t)b) 

Is this code valid for all the unsigned integer values of a & b?

Actually, I want unique result values for all possible unsigned integer values of a & b. The aim is to keep the size/length of the result minimal (in this case, we can bind it in 64 bit).

Answer 1

是的，它可以按您期望的方式工作（如果它们确实是未签名的）。

Answer 2

Another method:

uint32_t a = xxx;
uint32_t b = xxx;
uint64_t result;
uint32_t * p = (uint32_t *)&result;
p[0] = b;
p[1] = a;

Or maybe better:

union
{
  uint32_t b;
  uint32_t a;
  uint64_t res;
} u3264;
u3264 u;
u.a = xxx;
u.b = yyy;
// 64 bit result in u.res