C语言中的按位运算符：此函数如何工作？

Question

I can't figure out how this functions works:

void print3bytes(unsigned char s[],unsigned len)
{
    unsigned char *end;
    for(end=s+len-len%3;  s<end;  s=s+3 )
        printf("%x ",  s[0]|s[1]<<8  |  s[2]<<16); 
    putchar('\n'); 
}

how does this work for this example:

int main(void)
{
    unsigned char s[] = "\x12\x34\x56\x78\x9A\xBC\xDE\xFF";
    print3bytes(s, 8);
    return 0;
}

Could someone please explain?

Answer 1

The expression combines:

• The first byte s[0]
• The second byte s[1] shifted (`<<´) 8 bits to the left
• The third byte s[2] shifted ( << ) 16 bits to the left

using bitwise-or ( | ).

So, when s contains 0x12, 0x34, 0x56 as its first bytes, it computes

0x12 | (0x34 << 8) | (0x56 << 16)
  ^       ^             ^
  |       |             |
s[0]     s[1]         s[2]

which is

0x12 | 0x3400 | 0x560000

or 0x563412 . Basically it's unpacking (decoding, de-serializing, de-marshaling, whatever) a 24-bit integer in little-endian format from the start of s .

Answer 2

s[0] | s[1]<<8  |  s[2]<<16

Let's say s[2] is 0xff (255), s[1] is 0xaa (170) and s[0] is 0x11 (17). If you shift s[2] 16 bits, you get:

0xff0000

If you shift s[1] by 8bits, you get:

0x00aa00

And if you or these together you get

0xffaa00

And or this with s[0]:

0xffaa11