为什么-2147483648自动提升到可以适合int的长度？ [英] Why is -2147483648 automatically promoted to long when it can fit in int?

问题描述

#include <stdio.h>

int main()
{
    printf("%zu\n", sizeof(-2147483648));
    printf("%zu\n", sizeof(-2147483647-1));
    return 0;
}

上面的代码给出了输出（gcc）：

The above code gives as output (gcc):

8  
4  

为何<1c $ c> -2147483648 在1 ^st <$ c中自动提升为 long $ c> printf 即使它适合于 int ？

Why is -2147483648 automatically promoted to long in 1^st printf even when it can fit in an int?

另外，我在MinGW中也做了同样的处理，并输出结果：

Also, I tried the same in MinGW and it gives the output:

4  
4  

有人可以解释发生了什么吗？

Can someone please explain what's going on?

推荐答案

数字2147483648太大而无法放入 int 中，因此被升级至长。

The number 2147483648 is too large to fit into an int, so it is promoted to long.

然后，数字已被提升为 long ，计算其负数，产生-2147483648。

Then, after the number has already been promoted to long, its negative is computed, yielding -2147483648.

如果您好奇，可以看看 limits.h 。在我使用glibc的平台上，

If you're curious, you can look at limits.h. On my platform, which uses glibc,

#  define INT_MIN       (-INT_MAX - 1)
#  define INT_MAX       2147483647

在MinGW上， sizeof（long）== 4 ，所以升级到 long 不会削减它。根据C11标准，该值必须被提升为 long long 。这不会发生，可能是因为您的MinGW编译器默认为C90或更早版本。

On MinGW, sizeof(long) == 4, so promotion to long won't cut it. According to the C11 standard, the value must be promoted to long long. This doesn't happen, probably because your MinGW compiler is defaulting to C90 or earlier.

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