将unsigned char字节打包成c中的unsigned int

Question

我有一个赋值,我必须将4个unsigned char中的字节打包成unsigned int.

代码如下:

#include <stdio.h>

int main (){
    //Given this
    unsigned char a = 202; 
    unsigned char b = 254; 
    unsigned char c = 186; 
    unsigned char d = 190; 

    //Did this myself
    unsigned int u = a; 
    u <<=8; 
    u |= b; 
    u <<=8; 
    u |= c
    u <<=8; 
    U |= d; 
}

我知道:

u <<=8; 

将你的位向左移动8.但我对这些线条的作用感到困惑u |= b;？

简单地说,我试图更好地理解我所编写的代码将4个unsigned char中的字节打包成unsigned int.我以粗野的方式提出了这个解决方案.我只是试图以不同的方式打包字节,这种方式有效.但我不确定为什么.

先感谢您.

Answer 1

a这是202二进制的11001010

b这是254二进制的11111110

c这是186二进制的10111010

d这是190二进制的10111110

unsigned int u = a;
u <<= 8;    // now u would be 11001010 00000000
u |= b;     // now u would be 11001010 11111110
u <<= 8;    // now u would be 11001010 11111110 00000000
u |= c;     // now u would be 11001010 11111110 10111010
u <<= 8;    // now u would be 11001010 11111110 10111010 00000000
u |= d;     // now u would be 11001010 11111110 10111010 10111110
            // This is how        a        b        c        d    
            // are packed into one integer u.