Getting started #

Make sure to set-up first.

1. Write a function that prints a 32-bit integer in binary #

In C there is no way to use printf() such that it prints a variable, such as an int or unsigned int in its stored binary format. However, printing in binary comes in really handy sometimes. The goal is to write a function that takes an unsigned int x as input, and prints it in binary:

#include <stdio.h>

// <-- write print_binary(x) here

int main() {
    unsigned x = 27;
	print_binary(x); // 00000000 ... 00011011
    return 0;
}

To do so, first try to code a way to print a single bit of x. This requires the bitwise AND operator &, which is defined as: a & b returns a number that has a $1$ in a bit $j$ iff. both a and b have a $1$ in bit $j$. Otherwise the $j$-th bit is zero.

To print the $i$-th bit of $x$:

Make a number mask with zeros everywhere, but a $1$ in position $i$, using mask = 1 << i.
Apply the & between the mask and the input x.
Use the outcome to decide if bit $i$ is $0$ or $1$.

The solution is now to loop all the bits in x, and print them one by one.

Solution

One way to write this function is:

void print_binary(unsigned int x) {
    int mask;
    for (int i = 31; i >= 0; i--) {
        mask = 1 << i;
        if ((x & mask) == 0) {
            printf("0");
        } else {
            printf("1");
        }

        if (i % 8 == 0 && i > 0)
            printf(" ");
    }
    printf("\n");
}

The function works well with all 32-bit inputs, but the input has to be cast to 32-bit first.
It doesn’t work for any other types.

A compact way to write this function is:

void print_binary(unsigned x) {
    for (unsigned m = 1 << 31; m > 0; m >>= 1)
        printf("%d", (x & m) > 0)
}

Note:

mask >>= 1 is the same as mask = mask >> 1.
(x & mask) > 0 returns an integer (in C, logical operators, like >, result in an integer $0$ or $1$) and not in a boolean.