Decimal to binary function in Python using recursion-CodePudding

I am new in Python and trying to write a binary-to-decimal converted function like below

def decimaltobinary(n):
    if n > 1:
        decimaltobinary(n//2)
    print(n%2,end='')
    #return n%2

decimaltobinary(4)

This works perfectly fine. Now the question is when I am modifying it as below, it doesn't give me correct result -

def decimaltobinary(n):
    if n > 1:
        decimaltobinary(n//2)
    #print(n%2,end='')
    return n%2

a=decimaltobinary(4)
print(a)

Am I missing something with the return statement? Any pointer will be very helpful.

CodePudding user response：

In the second example returned value from decimaltobinary in if is completely ignored.

What you need to do is assign the returned value to a variable and then return it together with n%2.

Try this:

def decimaltobinary(n):
x = ''
if n > 1:
    x = decimaltobinary(n//2)
#print(n%2,end='')
return str(x)   ''   str(n%2)

a=decimaltobinary(4)
print(a)

CodePudding user response：

You need to be careful with your return statements. Try this:

def decimaltobinary(n):
    def _dtb(n, lst):
        if n <= 0:
            return lst
        lst.append(n&1)
        return _dtb(n>>1, lst)
    return ''.join(map(str, reversed(_dtb(n, list()))))
print(decimaltobinary(19))

Output: