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C Exercises: Determine the HCF of two numbers

C For Loop: Exercise-43 with Solution

Write a C program to find HCF (Highest Common Factor) of two numbers.

Pictorial Presentation:

Determine the HCF of two numbers

Sample Solution:

C Code:

#include <stdio.h>

void main()  
{  
    int i, n1, n2, j, hcf=1;  


     printf("\n\n  HCF of two numbers:\n ");
     printf("----------------------\n");


    printf("Input 1st number for HCF: ");  
    scanf("%d", &n1);  
    printf("Input 2nd number for HCF: ");  
    scanf("%d", &n2); 
  
    j = (n1<n2) ? n1 : n2;  
  
    for(i=1; i<=j; i++)  
    {  

        if(n1%i==0 && n2%i==0)  
        {  
            hcf = i;  
        }  
    }  
  
    printf("\nHCF of %d and %d is : %d\n\n", n1, n2, hcf);  
}

Sample Output: 

  HCF of two numbers:                                                                                         
 ----------------------                                                                                       
Input 1st number for HCF: 24                                                                                  
Input 2nd number for HCF: 28                                                                                  
                                                                                                              
HCF of 24 and 28 is : 4

Flowchart:

Flowchart : Determine the HCF of two numbers.

C Programming Code Editor:

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Previous: Write a program in C to convert a binary number into a decimal number without using array, function and while loop.
Next: Write a program in C to find LCM of any two numbers using HCF.

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C Programming: Tips of the Day

Printing hexadecimal characters in C:

You are seeing the ffffff because char is signed on your system. In C, vararg functions such as printf will promote all integers smaller than int to int. Since char is an integer (8-bit signed integer in your case), your chars are being promoted to int via sign-extension.

Since c0 and 80 have a leading 1-bit (and are negative as an 8-bit integer), they are being sign-extended while the others in your sample don't.

char    int
c0 -> ffffffc0
80 -> ffffff80
61 -> 00000061
Here's a solution: 
char ch = 0xC0;
printf("%x", ch & 0xff);

This will mask out the upper bits and keep only the lower 8 bits that you want.

Ref : https://bit.ly/3vOLizM