How to convert an int to a binary string represen

Question

I have an int that I want to store as a binary string representation. How can this be done?

Answer 1

Solution without reverse, no additional copy, and with 0-padding:

#include <iostream>
#include <string>

template <short WIDTH>
std::string binary( unsigned x )
{
    std::string buffer( WIDTH, '0' );
    char *p = &buffer[ WIDTH ];

    do {
        --p;
        if (x & 1) *p = '1';
    }
    while (x >>= 1);

    return buffer;
}

int main()
{
    std::cout << "'" << binary<32>(0xf0f0f0f0) << "'" << std::endl;
    return 0;
}

Answer 2

I have an int that I want to first convert to a binary number.

What exactly does that mean? There is no type "binary number". Well, an int is already represented in binary form internally unless you're using a very strange computer, but that's an implementation detail -- conceptually, it is just an integral number.

Each time you print a number to the screen, it must be converted to a string of characters. It just so happens that most I/O systems chose a decimal representation for this process so that humans have an easier time. But there is nothing inherently decimal about int.

Anyway, to generate a base b representation of an integral number x, simply follow this algorithm:

1. initialize s with the empty string

2. m = x % b

3. x = x / b

4. Convert m into a digit, d.

5. Append d on s.

6. If x is not zero, goto step 2.

7. Reverse s

Step 4 is easy if b <= 10 and your computer uses a character encoding where the digits 0-9 are contiguous, because then it's simply d = '0' + m. Otherwise, you need a lookup table.

Steps 5 and 7 can be simplified to append d on the left of s if you know ahead of time how much space you will need and start from the right end in the string.

In the case of b == 2 (e.g. binary representation), step 2 can be simplified to m = x & 1, and step 3 can be simplified to x = x >> 1.

Solution with reverse:

#include <string>
#include <algorithm>

std::string binary(unsigned x)
{
    std::string s;
    do
    {
        s.push_back('0' + (x & 1));
    } while (x >>= 1);
    std::reverse(s.begin(), s.end());
    return s;
}

Solution without reverse:

#include <string>

std::string binary(unsigned x)
{
    // Warning: this breaks for numbers with more than 64 bits
    char buffer[64];
    char* p = buffer + 64;
    do
    {
        *--p = '0' + (x & 1);
    } while (x >>= 1);
    return std::string(p, buffer + 64);
}

Answer 3

AND the number with 100000..., then 010000..., 0010000..., etc. Each time, if the result is 0, put a '0' in a char array, otherwise put a '1'.

int numberOfBits = sizeof(int) * 8;
char binary[numberOfBits + 1];
int decimal = 29;

for(int i = 0; i < numberOfBits; ++i) {
    if ((decimal & (0x80000000 >> i)) == 0) {
        binary[i] = '0';
    } else {
        binary[i] = '1';
    }
}
binary[numberOfBits] = '\0';
string binaryString(binary);