Please help me spot my error [closed] - c

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Hey I am writing a simple program that adds fractions.
but whenever i run the program it will not perform the operations.
It only scans the input, but no output.
please spot my error! >>>>>>CANNOT USE FLOATS<<<<<<<<
#include <stdio.h>
int main ( void )
{
int lcd, d1, d2, num1, num2, sum;
printf("Enter the first number:");
printf("Enter the numerator:");
scanf("%d", num1);
printf("Enter the denominator:");
scanf("%d", &d1);
printf("Enter the second number:");
printf("Enter the numerator:");
scanf("%d", &num2);
printf("Enter the denominator:");
scanf("%d", &d2);
for (lcd = d1; lcd % d2 != 0; lcd+= d1);
num1 *= lcd / d1;
num2 *= lcd / d2;
sum = num1 + num2;
printf("%d", sum);
return 0;
}

Answer 1

scanf must receive pointers
scanf("%d", num1) -> scanf("%d", &num1)
Because scanf use call by reference. What is that? please see the following code.
#include <stdio.h>
void foo(int i)
{
i = 1;
}
void bar(int *pi)
{
*pi = 1;
}
int main()
{
int a = 2;
foo(a);
printf("%d", a); /* output is 2 */
bar(&a);
printf("%d", a); /* output is 1 */
}
In foo(a), we're using call by value. That means a is copied when we call foo. the code, i = 1, of foo changes only the copy of a, and does NOT change real value of a.
In bar(&a), we're using call by reference. Can you find the difference? Yes, it's bar(&a), not bar(a). "&" operator gets the pointer of a, and we call bar with the pointer. So pi refers to a, and *pi = 1 changes real value of a successfully.
A real-life example is here: printf and scanf.
int i = 0;
printf("output number : %d\n", i);
scanf("%d", &i); /* input number */
printf doesn't need to change its arguments, so it use call-by-value.
But, scanf receives user's input and change its arguments into user's input. so it use call-by-reference. thus, we should use &i, not i.

Answer 2

Does it need to be that complex? Why not just divide the numerator by the denominator?
Also, your working with integers. I think you want floats.
#include <stdio.h>
int main ( void )
{
int d1, d2, num1, num2;
printf("Enter the first number:");
printf("Enter the numerator:");
scanf("%d", &num1);
printf("Enter the denominator:");
scanf("%d", &d1);
printf("Enter the second number:");
printf("Enter the numerator:");
scanf("%d", &num2);
printf("Enter the denominator:");
scanf("%d", &d2);
printf("%f",
((float) num1 / (float) d1) + ((float) num2 / (float) d2)
);
return 0;
}

Answer 3

If you want to compute the non simplified sum of two fractions, just using the following formula
a/b + c/d = (a*d + c*b)/(b*d)
Therefore, the following code could be used to achieve your goal
#include <stdio.h>
int main ( void )
{
int den1, den2, num1, num2, sum_num, sum_den;
printf("Enter the first number:\n");
printf("\tEnter the numerator:");
scanf("%d", &num1);
printf("\tEnter the denominator:");
scanf("%d", &den1);
printf("Enter the second number:\n");
printf("\tEnter the numerator:");
scanf("%d", &num2);
printf("\tEnter the denominator:");
scanf("%d", &den2);
sum_num = num1 * den2 + num2 * den1;
sum_den = den1 * den2;
printf("sum = %d/%d\n", sum_num, sum_den);
return 0;
}