什么是CHAR I = 0x80的，为什么溢出并没有发生位移 [英] what is char i=0x80 and why overflow did not happen in bit shifting

问题描述

下面是一个程序

#include <stdio.h>
main()
{ unsigned char i=0x80;
printf("i=%d",i<<1);
}

它给输出为256。
我不是什么不明确

The output it is giving is 256. I am not clear with what does

unsigned char i=0x80; <-- i is not int it is char so what will it store?

我知道bitshift和十六进制的东西。
如何i的值存储，它是如何被更改为256？

I know bitshift and hexadecimal things. How is the value of i being stored and how does it gets changed to 256?

更新

为什么当移位操作没有发生溢出没有发生？

Why did overflow not occurred when the bit shift operation happened?

推荐答案

在C，一个字符是用于存储字符数据的整数类型，通常为1个字节。

In C, a char is an integer type used to store character data, typically 1 byte.

保存在 i的值是 0x80的十六进制常数等于 128 。

The value stored in i is 0x80 a hexidecimal constant that is equal to 128.

在两个整数类型的算术运算（如 I＆LT;＆LT; 1 ）将提升到的更宽的类型，在这种情况下，到 INT ，因为 1 是一个整型常量。在任何情况下，整函数参数都提升到int类型。

An arithmetic operation on two integer types (such as i << 1) will promote to the wider type, in this case to int, since 1 is an int constant. In any case, integer function arguments are promoted to int.

然后将结果发送到的printf ，以％d个格式说明，这意味着打印整数

Then you send the result to printf, with a %d format specifier, which mean "print an integer".

这篇关于什么是CHAR I = 0x80的，为什么溢出并没有发生位移的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！