如何在不使用bitset的情况下显示unsigned int中的位？

Question

I am currently working on a project for school covering bit manipulation. We are supposed to show the bits for an unsigned integer variable and allow the user to manipulate them, turning them on and off and shifting them. I have all of the functionality working, except for displaying the bits once they have been manipulated. We are NOT allowed to use bitset to display the bits, and it will result in a heavy grade reduction.

I have tried using if statements to determine whether the bits are on or off, but this does not seem to be working. Whenever a bit is changed, it will simply print a lot of 0's and 1's.

std::cout << "Bits: ";

for (int i = sizeof(int)*8; i > 0; i--)
{

    if (a | (0 << i) == 1)
        std::cout << 1;
    if (a | (0 << i) == 0)
        std::cout << 0;

}

std::cout << std::endl << a;

I would expect that if I turn a bit on, that one bit will display a 1 instead of a 0 , with the rest of the bits being unchanged and still displaying 0 ; instead it prints a string of 1010101 about the length of half the console.

Answer 1

There are a couple of problems here, and you might want to do a detailed review of bit manipulation:

• for (int i = sizeof(int)*8; i > 0; i--) should be for (int i = sizeof(int)*8 - 1; i >= 0; i--) , because bits are 0-indexed (shifting 1 to the left 0 times gives a set bit on the rightmost position).
• We use bitwise AND ( & ) instead of bitwise OR ( | ) to check if a bit is set. This is because when we use bitwise AND with a number that only has a single bit set, the result will be a mask with the bit at the position of the 1 being in the same state as the corresponding bit in the original number (since anything AND 1 is itself), and all other bits being 0's (since anything AND 0 is 0).
• We want a mask with 1 in the position that we want to check and 0 elsewhere, so we need 1 << i instead of 0 << i .
• If the bit we're checking is set, we'll end up with a number that has one bit set, but that's not necessarily 1. So we should check if the result is not equal to 0 instead of checking if it's equal to 1.
• The == operator has a higher precedence compared to the | and the & operators, so parenthesis is needed.