Converting an integer base 10 to a binary number

[Robert Amesz]

François Granger wrote:

>Peter Stöhr <peter.stoehr at fh-hof.de> wrote:
>
>> Hi out there,
>> 
>> I know the int(x[,radix]) function that converts a string or a
>> number (for a give radix) to an integer. Is there a built-in
>> function that provides a conversion the other way round?
>> Something like
>>     bin(32) = "10000"
>
>Not fully polished, but works
>
># python
>"""
>create the binary rep of an integer as a string of '0' & '1'
>20010302    v 1.0
>"""
>def toBin(x):
>    a = ''
>    y = x
>    while x:
>        x, y = x / 2, y % 2
>        if y:
>            a = '1' + a
>        else:
>            a = '0' + a
>        y = x
>    return a
>
>if __name__ == '__main__'
>    l = [1, 2, 3, 4, 8, 16, 32, 64, 128, 255, 512, 1024, 2048, 4096,
>8192, 16384, 32768]
>    for x in l:
>        print toBin(x)
>


I'm afraid it doesn't work for x == 0, and you'll get into trouble when 
x is a float. (Strong typing does have its merits.) Also, for binary 
numbers we don't need the div-and-mod approach, a bitwise and (i.e. &) 
will do. (Incidentally, although I'm not using it here, there's a 
function called divmod() which is more efficient, presumably, than two 
seperate operations.) That leaves us with:


def toBin(x):
    x = int(x) # Just in case x is a float, 
    binstr = ''

    while 1: # Can't use while x, because x may be 0 initially
        if x & 1:
            binstr = '1' + binstr
        else:
            binstr = '0' + binstr
        x = x / 2
        if not x:
            return binstr


You can be even more concise (I don't know about the efficiency, 
though) by replacing the if/else statement by:

        binstr = string.digits[x & 1] + binstr
or:

        binstr = ('0', '1')[x & 1] + binstr


Robert Amesz