int2bin()

**dhuan** · 11-12-2010

This function returns a binary representation in string form, of a given integer.

Code:

#include <iostream>
#include <string>

using namespace std;

string int2bin(int num)
{
    string returnString;
    int pos=0;
    for(int x=0; x <= 15; x++)
    {
        pos = (pos == 0) ? 1 : pos + pos;
        returnString =  (num & pos) ? "1"+returnString : "0"+returnString;
    }
    return returnString;
}


int main()
{
    cout<<int2bin(255);
    return 0;
}

It seems to work fine when compiled;;;
I posted here in case you guys have any suggestions regarding better coding.

**NeonBlack** · 11-12-2010

You could do this with pos, which is a little simpler. (And I would probably name it mask instead of pos). Notice that I'm also reducing the scope of pos by declaring it inside the loop.

Code:

for (int i=0; i<16; ++i)
{
    int pos = 1<<i;
    ...
}

Another change you may want to consider is adding ones or zeros to the end of the string and then reversing it when you're done.

**iMalc** · 11-13-2010

You don't need x. Do the for-loop over pos, from 1 to 1<<16 and instead of increment it, you shift the mask left by one each time.
Prepending onto a string is not that efficient either, so I'd consider starting at 1<<15 and shifting right instead each time.

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