C program to efficiently multiply a number with 7 using bitwise operator

C program to efficiently multiply a number with 7 using bitwise operator

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  • Write a program in C to multiply a number with 7 using bitwise operator.
  • How to multiply a number by 7 in one line.

Required knowledge Bitwise operator in C

    • Let’s derive an expression to multiply a number with 7 using bitwise operator. Let N be the number that we want to multiply with 7.
    • N x 7 = N + N + N + N + N + N + N
    • N x 7 = N + N + N + N + N + N + N + (N – N)
    • N x 7 = (N + N + N + N + N + N + N + N) – N
    • N x 7 = 8xN – N
    • As we know that, left shifting any number by one bit multiply it by 2. Hence, multiplying any number with 8 is equivalent to right shifting it by 3 bits(For Example : NX3 = N << 3). Replacing 8xN in above statement by 8 << 3.

N x 7 = (N << 3) – N

WARNING !!!!
Multiply using bitwise operators: This approach can only be used to multiply integers or char by 7 because bitwise Operators can only be applied on char and integer operands. We cannot use bitwise operators with float, double, long double, void and other user define complex data types.

C program to multiply a number with 7 using bitwise operator.

C program to efficiently multiply a number with 7 using bitwise operator

#include<stdio.h>
 
int main() {
    int num;
     
    printf("Enter an integer\n");
    scanf("%d", &num);
     
    printf("7 X %d = %d", num, (num << 3) - num);
 
    return 0;
}

Output

Enter an integer
3
3 X 7 = 21

Enter an integer
0
0 X 7 = 0