CODEWITHSUNDEEP
cbitwise-operators
Sanjay-sopho
Sanjay-sopho

Reputation: 429

multiplying an integer by 5 using bitwise operator

I am trying to multiply an integer by 5 using bitwise operations. But there seems to be a bug in my code that I can't identify. Here's my code 

#include <stdio.h>

#define print_integer(exp) printf("%s : %d\n", #exp, (exp))

int multiply(int num) {
    int ans;
    ans = num << 2 + num;
    return ans;
}

int main() {
    int a = 1, b = 2, c = 3;
    print_integer(multiply(a));
    print_integer(multiply(b));
    print_integer(multiply(c));
    return 0;
}

Edit:- The bug is in line ans = num << 2 + num;

Upvotes: 3

Views: 1817

Answers (2)

chqrlie
chqrlie

Reputation: 144780

The precedence between << and + is counter intuitive. Use parentheses and compile with -Wall to get useful warnings abut this kind of potential mistake:

#include <stdio.h>

#define print_integer(exp) printf("%s : %d\n", #exp, (exp))

int multiply(int num) {
      return (num << 2) + num;
}

int main(void) {
    int a = 1, b = 2, c = 3;
    print_integer(multiply(a));
    print_integer(multiply(b));
    print_integer(multiply(c));
    return 0;
}

Upvotes: 3

Isaac Razilov
Isaac Razilov

Reputation: 19

You are adding the num after the change ( num<<2 ) , set a temp value for num . 

int tmp = num ; 
ans = num << 2 + tmp ;

Upvotes: -1

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