printf会导致未定义的行为吗？ [英] Can printf result in undefined behavior?

问题描述

int main() 
{
  unsigned int i = 12;
  printf("%lu", i); // This yields a compiler warning
}

在32位平台上， code> printf 使用％lu 结果是垃圾吗？

On a 32 bit platform, does using printf with an int using %lu result in garbage?

推荐答案

只有语句32位平台并不意味着 int 和 long 都有32位，以及它们的 unsigned 。

Only the statement "32 bit platform" doesn't mean that int and long both have 32 bits, as well as their unsigned counterparts.

如果 unsingned long ，为％lu 创建的时间长于 unsigned int 。

So yes, indeed this can happen if unsingned long, what %lu is made for, is longer than unsigned int.

但是即使长度相等，类型也不兼容，所以正式是未定义的行为。

But even if the lengths are equal, the types are not compatible, so formally it is undefined behaviour.

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