GCD function for C

Question

Q 1. Problem 5 (evenly divisible) I tried the brute force method but it took time, so I referred few sites and found this code:

#include<stdio.h>
int gcd(int a, int b)
{
    while (b != 0)
    {
        a %= b;
        a ^= b;
        b ^= a;
        a ^= b;
    }

    return a;
}

int lcm(int a, int b)
{
    return a / gcd(a, b) * b;
}

int main()
{
    int res = 1;
    int i;
    for (i = 2; i <= 20; i++)
    {
        res = lcm(res, i);
    }

    printf("%d\n", res);
    return 0;
}

This is very simple but I don't understand how function "gcd" works; can somebody please help me understand the logic. (I know it returns the GCD of 2 numbers but why so many operations?)

Answer 1

To your second question: The GCD function uses Euclid's Algorithm. It computes A mod B, then swaps A and B with an XOR swap. A more readable version might look like this:

int gcd(int a, int b)
{
    int temp;
    while (b != 0)
    {
        temp = a % b;

        a = b;
        b = temp;
    }
    return a;
}

Answer 2

This problem can also be solved in a very clean way with recursion:

int gcd(int a, int b) {
    int remainder = a % b;

    if (remainder == 0) {
        return b;
    }

    return gcd(b, remainder);
}

Answer 3

I executed this statements for GCD :

#include<stdio.h>
#include<conio.h>
int main(){
   int l, s,r;

   printf("\n\tEnter value : ");
   scanf("%d %d",&l,&s);

  while(l%s!=0){
    r=l%s;
    l=s;
    s=r;
  }
  printf("\n\tGCD = %d",s);
  getch();
}

Answer 4

The GCD computation in C :

int gcd(int a, int b){
    if (a && b) for(;(a %= b) && (b %= a););
    return a | b;
}

The absolute value computation :

#include <limits.h>
unsigned int abso(int v){
    const int mask = v >> (sizeof(int) * CHAR_BIT - 1);
    return (v + mask) ^ mask;
}

Answer 5

Using a bit of recursion and Objective-C

-(int)euclid:(int)numA numB:(int)numB
{
    if (numB == 0)
        return numA;
    else
        return ([self euclid:numB numB:numA % numB]);
}