是否保证将 -1 分配给无符号类型会产生最大值?

问题描述

我发现了一些关于这个特定主题的问题，但它们都是关于 C++ 的.

I found a few questions on this particular topic, but they were about C++.

将 -1 转换为无符号类型的可移植性如何?

将 -1 转换为无符号类型

将 -1 赋给 unsigned int 以获得最大值是否安全?

在阅读答案时，这似乎是 C 和 C++ 不同之处之一，这似乎很可能或至少不太可能.

And when reading the answers, it seemed likely or at least not unlikely that this is one of those things where C and C++ differs.

问题很简单:

如果我用 unsigned char/short/int/long var 声明一个变量或使用任何其他无符号类型，如固定宽度、最小宽度等，那么是否保证 var = -1 会将 var 设置为它可以容纳的最大值吗?该程序是否保证打印是"?

If I declare a variable with unsigned char/short/int/long var or use any other unsigned types like fixed width, minimum width etc, is it then guaranteed that var = -1 will set var to the maximum value it can hold? Is this program guaranteed to print "Yes"?

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

int main(void) {
    unsigned long var = -1;
    printf("%s\n", var == ULONG_MAX ? "Yes" : "No");
}

推荐答案

是的.

此程序是否保证打印是"?

是的.

这是一个转换，从 int -1 到 unsigned long.-1 不能表示为 unsigned long.来自 C11 6.3.1.3p2:

It's a conversion, from int -1 to unsigned long. -1 can't be represented as unsigned long. From C11 6.3.1.3p2:

否则，如果新类型是无符号的，则通过重复加或减一个新类型可以表示的最大值来转换该值，直到该值在新类型的范围内

所以我们将一个 (ULONG_MAX + 1) 添加到 -1，我们得到 -1 + (ULONG_MAX + 1) = ULONG_MAX位于 unsigned long 的范围内.

so we add one (ULONG_MAX + 1) to -1 and we get -1 + (ULONG_MAX + 1) = ULONG_MAX which is in range of unsigned long.

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