printf 64位类型说明符问题 [英] printf 64bit type specifier issue

问题描述

我在 msdev 2010 中看到 __int64 类型的一些奇怪行为.有人能告诉我发生了什么吗?我想这里有两个问题，首先是如何显示 64 位整数，其次是行为 - 即为什么看起来 __int64 实际上是 32 位整数...

I'm seeing some strange behaviour with __int64 types with msdev 2010. Could anyone enlighten me as to what is going on ? I guess there are 2 issues here, firstly how to display 64bit ints, secondly the behaviour - i.e. why is it looking like __int64 is actually a 32bit int...

#include <stdio.h>

int main()
{
  int vl_idx;
  unsigned __int64 vl_64;
  unsigned __int64 vl_64_test;

  for (vl_idx = 0; vl_idx < 64; vl_idx++)
  {
    vl_64 = 1 << vl_idx;
    printf ("vl_64 (%d) = %I64u\n", vl_idx, vl_64);
    printf ("vl_64 (%d) = %llu\n", vl_idx, vl_64);
    printf ("vl_64 (%d) = %lu\n", vl_idx, vl_64);
  }
  vl_64_test = 1 << 31;
  if (vl_64 > vl_64_test)
     printf ("greater\n");
  if (vl_64 == vl_64_test)
     printf ("equal\n");
  if (vl_64 < vl_64_test)
     printf ("less\n");

  return 0;
}

前 30 次迭代的输出符合预期:

The output is as expected for the first 30 iterations:

vl_64 (0) = 1
vl_64 (0) = 1
vl_64 (0) = 1
vl_64 (1) = 2
vl_64 (1) = 2
vl_64 (1) = 2
...
vl_64 (30) = 1073741824
vl_64 (30) = 1073741824
vl_64 (30) = 1073741824
vl_64 (31) = 18446744071562067968
vl_64 (31) = 18446744071562067968
vl_64 (31) = 2147483648
vl_64 (32) = 1
vl_64 (32) = 1
vl_64 (32) = 1
vl_64 (33) = 2
vl_64 (33) = 2
vl_64 (33) = 2
...
vl_64 (62) = 1073741824
vl_64 (62) = 1073741824
vl_64 (62) = 1073741824
vl_64 (63) = 18446744071562067968
vl_64 (63) = 18446744071562067968
vl_64 (63) = 2147483648
equal

但后来事情?溢出?在第 32 次迭代.这可能只是显示问题，但最后的比较表明并非如此.这是使用 msdev 2010 cl(64 位版本)编译并在 64 位 Windows 操作系统(具有 64 位 CPU)上运行.关于为什么比较指出 1<<31 == 1<<63 的任何建议?

But then things ?overflow? at the 32nd iteration. It's possible this is just a display issue, but the comparison at the end suggests otherwise. This was being compiled with the msdev 2010 cl (64 bit version) and running on a 64bit windows OS (with a 64 bit CPU). Any suggestions as to why the comparison states that 1<<31 == 1<<63 ?

感谢您的建议，

吉姆

推荐答案

在处理比 int 更宽的任何内容时，您需要小心整数文字，例如你需要改变:

You need to be careful with integer literals when dealing with anything wider than int, e.g. you need to change:

vl_64 = 1 << vl_idx;

到:

vl_64 = 1LLU << vl_idx;

否则，在被隐式转换为无符号 64 位结果之前，首先将右侧评估为 int 表达式.

otherwise the right hand side is evaluated as an int expression first before being implicitly cast to an unsigned 64 bit result.

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