Integer to binary stored in an array

Question

TL:DR I have a 10-bit integer and want it as a binary number stored in an array.

Let's say I have an int from 0-1023 and I want that number converted to a 10-bit binary number that I will then put in an boolean array. Here is what I currently do.

void toBinary(bool *binary)
{
    string binaryStr = bitset<10>(myValue).to_string(); //to binary
    for (int i = 0; i < 10; i++) binary[i] = binaryStr[i] - '0';
}

Is there any better way then this? I will be doing some video processing with this and I'll be running it on a Pi, so I need my program to be as light and fast as possible.

Why are you converting to a string? Why not go directly from the bitset to the bool buffer? — PaulMcKenzie
– PaulMcKenzie, Commented Jan 20, 2017 at 1:42
@PaulMcKenzie : It's the only way I know of. The .to_ulong didn't really worked well for me. That's what I want to get rid of. — Dat Ha
– Dat Ha, Commented Jan 20, 2017 at 1:43
Maybe you missed the fact that a bitset<10> can be stored in a variable, and then use the variable in a loop? — PaulMcKenzie
– PaulMcKenzie, Commented Jan 20, 2017 at 1:49

Tomáš Zato · Accepted Answer · 2017-01-20 01:54:37Z

I don't understand why do you convert the bitset to string. Just access it's elements directly.

bitset<10> myBitset(myValue);
for (int i = 0; i < 10; i++)
    binary[i] = myBitset[i];

You could also go with bit shifting, that's the most low level way of doing this:

int mask = 1; // binary 10000000 00000000 ...
for (int i = 0, l = NUM_BITS; i < l; ++i) {
    // binary & operation does 
    // AND logic operation for all corresponging bit
    // so 0010&0011=0010
    binary[i] = myValue & mask;
    // move the bits in mask one to the right
    mask = mask>>1;
}

If you're going with bitset, I'd recommend that you keep the bitset instead of using bool* because in bitsets, every bit actually occupies one bit, whereas bool is eight bits large at least.

Finally, here's some test code I made for that, you can use it to do benchmarks:

#include <iostream>
#include <bitset>
#define NUM_BITS 10
int main(int argc, char *argv[])
{
    const int numBits = NUM_BITS;
    bool binary[numBits];
    const int myValue = 1;

    std::bitset<NUM_BITS> myBitset(myValue);
    //for (int i = 0; i < NUM_BITS; i++)
        //binary[i] = myBitset[i];

    for (int i = 0, l = NUM_BITS; i < l; ++i) {
        std::cout<< (binary[i]?'1':'0')<<" ";
    }
    std::cout<<"\n";
    int mask = 1; // binary 10000000 00000000 ...
    for (int i = 0, l = NUM_BITS; i < l; ++i) {
        // binary & operation does 
        // AND logic operation for all corresponging bit
        // so 0010&0011=0010
        binary[i] = myValue & mask;
        // move the bits in mask one to the right
        mask = mask>>1;
    }

    for (int i = 0, l = NUM_BITS; i < l; ++i) {
        std::cout<< (binary[i]?'1':'0')<<" ";
    }
}

I am using a bool because I will need to use them as my GPIO states. Thanks for your answer though :)

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