在无法到达的代码中停止GCC换档警告

Question

Given:

#include <stdio.h>
#include <limits.h>

int main()
{
  if (sizeof (long) > sizeof (int)) {
    long x = 42;
    x <<= CHAR_BIT * sizeof (int);
  }

  printf("sizeof (long) == %d\n", (int) sizeof (long));
  printf("sizeof (int) == %d\n", (int) sizeof (int));

  return 0;
}

On a platform where the sizes are equal I get this, with various version of GCC:

$ gcc -Wall shiftcomplain.c  -o shiftcomplain
shiftcomplain.c: In function ‘main’:
shiftcomplain.c:8:5: warning: left shift count >= width of type [enabled by default]
$ ./shiftcomplain 
sizeof (long) == 4
sizeof (int) == 4

The code block is not reachable when the types have an equal size, so the bad shift will never execute. It will only execute if long is wider than int , in which case the shift will not be out of range for that type.

How can we eliminate this annoying warning, with these constraints:

• I don't want to disable it globally because it is useful (when not falsely positive).

• I don't want split the shift---that is, perform it as two consecutive left shifts which add to the desired shift size.

• I don't want to convert the if test into a preprocessor #if . (This is easy to do in this case with INT_MAX and LONG_MAX , but troublesome in the actual program.)

---

Based on nm 's answer, I'm using something very similar to the following pattern:

const int condition = sizeof (long) > sizeof (int);

if (condition) {
  /*...*/
  x <<= CHAR_BIT * sizeof (int) * condition;
}

This pattern applied in my actual code suppresses the diagnostic, and the generated code doesn't change compared to not multiplying by condition .

Answer 1

x <<= (sizeof (long) > sizeof (int) ? CHAR_BIT * sizeof (int) : 0);

Answer 2

我与nm处于同一条路径，但提出了以下内容，从语义上看，这可能是预期的内容（最大sizeof（int）个字节来自x）。

x <<= (sizeof(long) - sizeof(int))*CHAR_BIT;