multiplication_with_shift_operator_v1.c

/*

    multiplication_with_shift_operator_v1.c

    https://www.geeksforgeeks.org/multiplication-two-numbers-shift-operator/
    https://www.geeksforgeeks.org/left-shift-right-shift-operators-c-cpp/

    How the C/C++ << and >> Shift Operators Work
    https://www.eeweb.com/profile/max-maxfield/articles/how-the-c-c-shift-operators-work

    Left shifting an integer “x” with an integer “y” (x<<y)
    is equivalent to multiplying x with 2^y (2 raise to power y).

    Similarly right shifting (x>>y) is equivalent to dividing x with 2^y.

    The Joyful Programmer:
Old optimization statement that I learned many, many, many
years ago, when reading an optimization book by Michael Abrash. It goes:

"Multiply if you must, but never divide."

Remove the multiplication operator. Why? Well,
when C++ is reduced to machine language, it takes the processor (newer ones)
about 30 clock cycles to execute a single multiplication. Division takes over
60 clock cycles. However, we can reduce the amount of clock cycles by using a
shift operator.

That's only 2 clock cycles versus an estimated 30 clock cycles, a 28 clock
cycle difference, which could be essential if you are doing millions of
calculations.


    There are also opposing opinions:
    Is multiplication and division using shift operators in C actually faster?
    This depends on the processor and the compiler. Some compilers already
    optimize code this way, others don't. So you need to check each time
    your code needs to be optimized this way.
    https://stackoverflow.com/questions/6357038/is-multiplication-and-division-using-shift-operators-in-c-actually-faster


    You can find all my C programs at Dragan Milicev's pastebin:

    https://pastebin.com/u/dmilicev

*/

#include <stdio.h>

int multiply(int n, int m)
{
    int ans = 0, count = 0;
    while (m)
    {
        if (m % 2 == 1)         // check for set bit and left
            ans += n << count;  // shift n, count times

        count++;                // increment of place value (count)
        m /= 2;
    }
    return ans;
}

int main(void)
{
    int n = 20 , m = 13;

    printf("\n %d * %d = %d \n", n, m, multiply(n, m) );


    return 0;

} // main()