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u |= b;do这些线条感到困惑？

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是二进制的202将是11001010

b是二进制的254将是11111110

c是二进制的186将是10111010

d是二进制的190将是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.

Answer 2

u | = b表示u = u OR b因此，这是OR运算