Execution bug in Gcd Program - c

Question

I am trying to write a program for gcd(Greatest Common Divisor),I know the code is really long and inefficient but still it should run. It is taking the input of numbers but not executing the final printf statement and not doing the processing of decision making blocks. Can u tell me what am I doing wrong here?
#include <stdio.h>
void main()
{
int a, b;
printf("Enter the nos for gcd:");
scanf("%d\n%d", &a, &b);
int temp;
if (a > b)
{
for (int i = 0; i <= b; i++)
{
if ((a % i == 0) && (b % i == 0))
{
temp = i;
}
}
}
else if (a == b)
{
for (int i = 0; i <= a; i++)
{
if ((a % i == 0) && (b % i == 0))
{
temp = i;
}
}
}
else if (b > a)
{
for (int i = 0; i <= a; i++)
{
if ((a % i == 0) && (b % i == 0))
{
temp = i;
}
}
}
printf("Gcd for %d and %d is:%d", a, b, temp);
}
I tried making program for gcd and` was expecting the same

Answer 1

After reviewing your code and the good comments above, I decided it might be beneficial to you to distill these comments into what I found in debugging your program and providing how these good comments could be illustrated in a refactored version of your code for your edification.
First off, when I compiled and tested your program, I received a segmentation fault. This was due to the issue with your "for" loop values starting with a value of zero.
for (int i = 0; i <= b; i++)
That, coupled with attempting to find the modulo value of zero causes a "divide by zero" situation, which is not defined.
Revising each "for" loop from "int i = 0" to "int i = 1" gets one past the segmentation fault and allows the program to function.
#Vera:~/C_Programs/Console/GCDenominator/bin/Release$ ./GCDenominator
Enter the nos for gcd:45
60
Gcd for 45 and 60 is:15
The other thing I observed and was noted by the commentators is the fact that the program, as is, is repeating the same essential "for" loop three times for conditions of "a > b", "a == b", and "b > a". Seeing basically redundant code like that should be a signal that there could be a better way to streamline the use of "for" loops to possibly one iteration of the "for" loop. With that in mind, following is a refactored version of your program.
#include <stdio.h>
int main()
{
int a, b, c, temp;
printf("Enter the nos for gcd:");
scanf("%d\n%d", &a, &b);
if (a > b) /* Place the greater value into a work variable to simplify the loop test */
{
c = a;
}
else
{
c = b;
}
for (int i = 1; i <= c; i++) /* Start at "1" - otherwise a program dump can occur */
{
if ((a % i == 0) && (b % i == 0))
{
temp = i;
}
}
printf("GCD for %d and %d is:%d\n", a, b, temp);
return 0; /* Customary for the main function to return an integer */
}
Some points to note.
As noted in the comments above, you had defined the main function to not return a value (void); however, it is usual and customary for the main function to return an integer value, so the main function was redefined as such.
An additional work variable is added and used to contain the greater of the first or second value entered and then used in a single "for" loop to streamline the testing of greatest common denominator.
As noted earlier, the initial value for the "for" loop is "1" rather than "0" to avoid the segmentation fault being received when running the initial version of the program.
Following were some sample tests of the revised program.
#Vera:~/C_Programs/Console/GCDenominator/bin/Release$ ./GCDenominator
Enter the nos for gcd:45
60
GCD for 45 and 60 is:15
#Vera:~/C_Programs/Console/GCDenominator/bin/Release$ ./GCDenominator
Enter the nos for gcd:60
45
GCD for 60 and 45 is:15
#Vera:~/C_Programs/Console/GCDenominator/bin/Release$ ./GCDenominator
Enter the nos for gcd:144
144
GCD for 144 and 144 is:144
#Vera:~/C_Programs/Console/GCDenominator/bin/Release$ ./GCDenominator
Enter the nos for gcd:43
67
GCD for 43 and 67 is:1
Again, review this possible solution along with the helpful comments provided as you learn C programming. And, it probably would be a great idea to acquire some tutorial books on the C programming language as they usually have solid information on the language along with proven coding techniques.
Try this out and see if it meets the spirit of your project.