C++ cout函数打印

Question

Why this function:

uint64_t rot_xor(uint64_t a1, uint64_t a2) {
    int size = sizeof(a1)*4;
    cout<<" "<<"size:"<<bitset<8>(size).to_string()<<" "<<size;
    int shift;
    uint64_t output = 0;
    cout<<endl;
    for (shift = 0; shift < size; shift++)
        if(a1&(1<<shift)) {
            output ^= (a2 << shift) | (a2 >> (size - shift));
            cout << bitset<64>(output).to_string()<<endl;
        }
return output;
}

Print output:

---

1. : size:00010000 20
2. : 0000000000000000000000000000000010110000110000011111001011111001

Answer 1

point 1

If the question is why your program prints 20 as a result of what seems to be sizeof(uint64_t) * 4 instead of 32, well it's because a few lines before calling rot_xor there is a:

cout << hex

In fact, the exact output as can be seen in the link you posted is

size:00100000 20

or 32 in binary and hexadecimal representation.

point 2

I have no idea what your expected ouput should be like.