打印值0x89（-119）时发生奇怪的输出

问题描述

正如标题所示，运行以下代码时出现怪异结果：

  #include  $ b int main（）
 {
 char buff [4] = {0x17，0x89，0x39，0x40}; 
 unsigned int * ptr =（unsigned int *）buff; （3 * 8））; 
 char a =（char）（（* ptr <>（3 * 8））; 
 
 printf（0x％x \ n，* ptr）; 
 printf（0x％x \，a）; 
 printf（0x％x \ n，b）; 
 printf（0x％x \，c）; 
 printf（0x％x \ n，d）; 
 
返回0; 
} 
  

输出：

  0x40398917 
 0x40 
 0x39 
 0xffffff89 
 0x17 
  

为什么我没有得到 0x89 ？

解决方案这是因为你的 char 变量是被签名的，并且它们在被升级（在这种情况下被升级到更宽的类型）时正在进行符号扩展。签名扩展是在进行此促销时保留标志的一种方式，因此 -119 保持为 -119 ，无论它是8位，16位或更宽的类型。

您可以通过明确使用 unsigned char 来修复它，因为，至少在C中， char 是有符号还是无符号是特定于实现的。从 C11 6.2.5类型/ 15 ：

签名扩展名无法用于无符号类型因为他们，...好，没有签名： - ）

As the title says, I get a "weird" result when running the following code:

#include <stdio.h>

int main()
{
    char buff[4] = {0x17, 0x89, 0x39, 0x40};
    unsigned int* ptr = (unsigned int*)buff;
    char a = (char)((*ptr << (0*8)) >> (3*8));
    char b = (char)((*ptr << (1*8)) >> (3*8));
    char c = (char)((*ptr << (2*8)) >> (3*8));
    char d = (char)((*ptr << (3*8)) >> (3*8));

    printf("0x%x\n", *ptr);
    printf("0x%x\n", a);
    printf("0x%x\n", b);
    printf("0x%x\n", c);
    printf("0x%x\n", d);

    return 0;
}

Output:

0x40398917
0x40
0x39
0xffffff89
0x17

Why am I not getting 0x89 ?

解决方案

It's because your char variables are signed and they're undergoing sign extension when being promoted (upgraded to a wider type in this case). Sign extension is a way of preserving the sign when doing this promotion, so that -119 stays as -119 whether it's 8-bit, 16-bit or a wider type.

You can fix it by explicitly using unsigned char since, in C at least, whether char is signed or unsigned is implementation-specific. From C11 6.2.5 Types /15:

Sign extension does not come into play for unsigned types because they're, ... well, unsigned :-)

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