c++ - gcc:警告:大整数隐式截断为无符号类型

Question

#include<stdio.h>

int main()
{

    unsigned char c;
    c = 300;
    printf("%d",c);
    return 0;
}

输出是可预测的还是未定义的？

Answer 1

很抱歉第一个回答，这里是 C++ 标准的解释:)

Is the output in any way predictable or its undefined??

这是可以预见的。这段代码有两点需要注意:一、unsigned char类型不能赋值:

unsigned char c;
c = 300;

3.9.1 Fundamental types (Page 54)

Unsigned integers, declared unsigned, shall obey the laws of arithmetic modulo 2n where n is the number of bits in the value representation of that particular size of integer.41)
...
41) This implies that unsigned arithmetic does not overflow because a result that cannot be represented by the resulting unsigned integer type is reduced modulo the number that is one greater than the largest value that can be represented by the resulting unsigned integer type.

基本上:

c = 300 % (std::numeric_limits<unsigned char>::max() + 1);

其次，在printf的格式字符串中传递%d来打印unsigned char变量。
这个ysth做对了 ;) 没有未定义的行为，因为在 variadic arguments 的情况下会发生从 unsigned char 到 int 的提升转换!

注意:答案的第二部分是对 the comments of this answer 中所说内容的改写但这本来不是我的答案。