C ++说明将uint64强制转换为uint32

Question

I'm trying to cast a uint64_t (representing time in nanoseconds from D-day using a boost chrono high precision clock) to a uint32_t in order to seed a random number generator.

I just want the least significant 32 bits of the uint64_t. Here is my attempt:

uint64_t ticks64 = dtn.count(); // This has the ticks in nanosec
uint64_t ticks32_manual = ticks64 & 0xFFFFFFFF;
uint32_t ticks32_auto = (uint32_t) ticks64;
mexPrintf("Periods: %llu\n", ticks64);
mexPrintf("32-bit manual truncation: %llu\n", ticks32_manual);
mexPrintf("32-bit automatic truncation: %u\n", ticks32_auto);

The output of my code is as follows:

Periods: 651444791362198

32-bit manual truncation: 1331774102

32-bit automatic truncation: 1331774102

I was expecting the last few digits of the 32 and original 64-bit representations to be the same, but they are not. That is, I thought I would "lose the left half" of the 64-bit number.

Can anyone explain what's going on here? Thanks.

Btw, I've seen this link .

Answer 1

As pointed out in the comments there's nothing wrong with the operation of your code, it's just that you're not visualizing the output correctly. Here's your code, corrected and runnable:

#include <cstdio>
#include <cstdint>

int main() {
  uint64_t ticks64 = 651444791362198llu;
  uint64_t ticks32_manual = ticks64 & 0xFFFFFFFF;
  uint32_t ticks32_auto = (uint32_t) ticks64;

  printf("Periods: %llX\n", ticks64);
  printf("32-bit manual truncation: %llX\n", ticks32_manual);
  printf("32-bit automatic truncation: %X\n", ticks32_auto);
}

And the output is:

Periods: 2507C4F614296
32-bit manual truncation: 4F614296
32-bit automatic truncation: 4F614296