Convert an integer to binary without using the built-in bin function

Question

This function receives as a parameter an integer and should return a list representing the same value expressed in binary as a list of bits, where the first element in the l

Answer 1

Not really the most efficient but at least it provides a simple conceptual way of understanding it...

1) Floor divide all the numbers by two repeatedly until you reach 1

2) Going in reverse order, create bits of this array of numbers, if it is even, append a 0 if it is odd append a 1.

Here's the literal implementation of that:

def intToBin(n):
    nums = [n]
    while n > 1:
        n = n // 2
        nums.append(n)

    bits = []
    for i in nums:
        bits.append(str(0 if i%2 == 0 else 1))
    bits.reverse()
    print ''.join(bits)

Here's a version that better utilizes memory:

def intToBin(n):
    bits = []

    bits.append(str(0 if n%2 == 0 else 1))
    while n > 1:
        n = n // 2
        bits.append(str(0 if n%2 == 0 else 1))

    bits.reverse()
    return ''.join(bits)

Answer 2

may I propose this:

def tobin(x,s):
    return [(x>>k)&1 for k in range(0,s)]

it is probably the fastest way and it seems pretty clear to me. bin way is too slow when performance matters.

cheers

Answer 3

Convert integer to list of bits with a fixed length :

[int(x) for x in list('{0:0{width}b}'.format(8, width=5))]