将unsigned char字节打包到c中的unsigned int中

Question

I have an assignment in which I have to pack the bytes from 4 unsigned char into an unsigned int.

the code goes as following:

#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; 
}

I know that:

u <<=8; 

Shifts the bits in u to the left 8. But I am confused as to what the lines like u |= b; do?

Simply, I am trying to better understand what the code I came up works into packing the bytes from 4 unsigned char into an unsigned int. I came up with this solution in a brute type of way. I was just trying to pack bytes in different ways, and this way worked. But I am not really sure why.

Thank you in advance.

Answer 1

a which is 202 in binary would be 11001010

b which is 254 in binary would be 11111110

c which is 186 in binary would be 10111010

d which is 190 in binary would be 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