添加两个无符号char变量，结果为int [英] Adding two unsigned char variables and result is int

问题描述

有代码:

#include <iostream>
int main(){
  unsigned char a = 4, b = 255;
  int g = (unsigned char)a + (unsigned char)b;
  std::cout << g << std::endl;
  return 0;
}

结果:

259

为什么结果是259，而不是3?如果添加了两个无符号的char变量，则应该有溢出，结果应为3，然后应将其从无符号的char 3转换为int 3.

Why the result is 259, not 3? If there are added two unsigned char variables, there should be overflow, result should be 3 and then it should convert from unsigned char 3 to int 3.

推荐答案

添加操作将首先推广在进行加法之前，将操作数添加到 int .这就是C的工作方式.如果要截断，则需要将其分配回较窄的类型，例如 unsigned char .

The addition operation will first promote its operands to int, before doing the addition. This is how C works. If you want to truncate, you need to assign it back into a narrower type, such as unsigned char.

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