Compare Bits in C

This article will explain several methods of how to compare bits in C.

Implement a Custom Function using Bitwise XOR and AND Operations for Bit Comparison in C

Generally, bit comparison entails accessing single bit values and conducting needed operations, such as implementing bit-field using union and struct keywords. Although, bitwise operations offer a more efficient method of comparing specified bits in numbers. In this case, we implement a separate function suited for the u_int32_t type, which is guaranteed to have 32-bit width.

The example program takes three integers as command-line arguments, the first two of which are the numbers that are compared, while the third integer specifies the n-th bit. Notice that we convert argv elements using strtol, thus losing some precision when storing the return values in u_int32_t type. The compareBits function declares two local variables storing intermediate values - mask and tmp. We set the n-th bit in the mask by shifting the 1 left by nth - 1 positions. Then, both user input numbers are XOR-ed to get the bit differences in each position with a set bit in result denoting different values. Finally, we need to extract the bit from the n-th position and check if the value is 0.

#include <stdlib.h>
#include <stdio.h>
#include <string.h>
#include <stdbool.h>

bool compareBits(u_int32_t n1, u_int32_t n2, u_int32_t nth) {
    u_int32_t mask, tmp;

    mask = 1 << (nth - 1);
    tmp = n1 ^ n2;

    if ((tmp & mask) == 0)
        return true;
    else
        return false;
}

int main(int argc, char *argv[]) {
    u_int32_t num1, num2, bit;

    if (argc != 4) {
        fprintf(stderr, "Usage: %s integer1 integer2 nth_bit \n", argv[0]);
        exit(EXIT_FAILURE);
    }

    num1 = strtol(argv[1], NULL, 0);
    num2 = strtol(argv[2], NULL, 0);
    bit = strtol(argv[3], NULL, 0);

    compareBits(num1, num2, bit) ? printf("bits equal!\n") : printf("bits not equal!\n");

    exit(EXIT_SUCCESS);
}

sample File Format:

./program 1234 1231 1

Input File Format:

bits not equal!

As an example, the integers 1234 and 1231 are represented in binary as 00000000000000000000010011010010 and 00000000000000000000010011001111, respectively. Thus, XOR-ing these two result in 00000000000000000000000000011101 binary representation, which is finally AND-ed with the mask 00000000000000000000000000000010 to extract the single bit position value. If the result is all zeros, it implies that compared bits are equal; otherwise, it is the opposite.

Alternatively, though, we can shorten the compareBits function and move XOR/AND operations to the return statement as shown in the following example code. Note that the function returns the logical opposite of bitwise XOR/AND operations, as we output the corresponding messages in the caller function using the ? : tenary statement.

#include <stdlib.h>
#include <stdio.h>
#include <string.h>
#include <stdbool.h>

bool compareBits(u_int32_t n1, u_int32_t n2, u_int32_t nth) {
    u_int32_t mask;

    mask = 1 << (nth - 1);

    return !((n1 ^ n2) & mask);
}

int main(int argc, char *argv[]) {
    u_int32_t num1, num2, bit;

    if (argc != 4) {
        fprintf(stderr, "Usage: %s integer1 integer2 nth_bit \n", argv[0]);
        exit(EXIT_FAILURE);
    }

    num1 = strtol(argv[1], NULL, 0);
    num2 = strtol(argv[2], NULL, 0);
    bit = strtol(argv[3], NULL, 0);

    compareBits(num1, num2, bit) ? printf("bits equal!\n") : printf("bits not equal!\n");

    exit(EXIT_SUCCESS);
}

sample File Format:

./program 1234 1231 2

Input File Format:

bits equal!