What will be in memory uint64_t(6) | (uint64_t('V') << 8) | (uint64_t('E') << 16) | (uint64_t('S') << 24) | (uint64_t('T') << 32) | (uint64_t('S') << 40)

Kofr

I can not explain why this bitwise operation results following bitset in memory.

#include <iostream>
#include <bitset>

int main(int , char **)
{   
    std::bitset<64> bitRepresentation {
        uint64_t(6) | (uint64_t('V') << 8) | (uint64_t('E') << 16) | (uint64_t('S') << 24) | (uint64_t('T') << 32) | (uint64_t('S') << 40)
         };
    std::cout << bitRepresentation;
    return 0;
}

where
V = 56 = 111000
E = 45 = 101101
S = 53 = 110101
T = 54 = 110110

the result
0000000000000000010100110101010001010011010001010101011000000110

I can understand why 110 to the right of bitset is 6 but other bit codes are weird. Why is other digits are they are?

Paul Colby

Hi @Kofr,

It looks right to me.

0000000  00000000  01010011  01010100  01010011  01000101  01010110  00000110
   -         -         S         T         S         E         V       int(6)

The first 16 bits are all 0, because you set nothing in those bits. The next 5 bytes are ASCII chars STSEV, and the final byte is integer 6.

V = 56 = 111000

Actually, V = 0x56 = 86 (decimal) = 01010110.

Cheers.