Question

I thought shift operator shifts the memory of the integer or the char on which it is applied but the output of the following code came a surprise to me.

#include <stdio.h>
#include <stdlib.h>
#include <stdint.h>

int main(void) {
    uint64_t number = 33550336;

    unsigned char *p = (unsigned char *)&number;
    size_t i;
    for (i=0; i < sizeof number; ++i)
        printf("%02x ", p[i]);
    printf("\n");

    //shift operation
    number = number<<4;

    p = (unsigned char *)&number;
    for (i=0; i < sizeof number; ++i)
        printf("%02x ", p[i]);
    printf("\n");

    return 0;
}

The system on which it ran is little endian and produced the following output:

00 f0 ff 01 00 00 00 00 
00 00 ff 1f 00 00 00 00 

Can somebody provide some reference to the detailed working of the shift operators?

Answer 1

I think you’ve answered your own question. The machine is little endian, which means the bytes are stored in memory with the least significant byte to the left. So your memory represents:

00 f0 ff 01 00 00 00 00 => 0x0000000001fff000
00 00 ff 1f 00 00 00 00 => 0x000000001fff0000

As you can see, the second is the same as the first value, shifted left by 4 bits.