Definitely less elegant than other solutions, but for my fellow beginners:

def flippingBits(n):

        # turn 'n' into binary string
        l=format(n, '032b')
        m=[]

        # flip all bits
        for i in range(len(l)):
                m.append(1-int(l[i]))

        num=0
        if m[-1]==1:
                num+=1
        for i in range(len(m)-1):

                # start from the back of the string, skip m [ -1 ]
                curr=int(m[-i-2])
                num+=(2*curr)**(i+1)
        return(num)

In Python 3:

def flippingBits(n):
    x = int('0b11111111111111111111111111111111', base=0)
    return x^n

in python:

def flippingBits(n): return 2**32-n-1

Is long as input/output in the C version expected? It should probably be uint32_t to match spec.

JAVA solution:

public static long flippingBits(long n) {
        return ~n & 0xFFFFFFFFL;
    }