Trouble of pointers as function arguments - c

Question

I have created a function that does the euclidean division in C and putting the quotient and remainder as function arguments.
code listing 1:
#include<stdio.h>
#include<stdlib.h>
void euclidean(int a, int b, int * q, int * r);
int main(int argc, char ** argv){
int * q;
int * r;
euclidean(5,4, q, r);
printf("q = %d, r = %d", *q, *r);
return 0;
}
void euclidean(int a, int b, int * q, int * r){
q = (int*)malloc(sizeof(int));
r = (int*)malloc(sizeof(int));
*q = a/b;
*r = a % b;
//printf("q = %d, r = %d", *q, *r); //this will show
return;
}
code listing 2:
#include<stdio.h>
#include<stdlib.h>
void euclidean(int a, int b, int * q, int * r);
int main(int argc, char ** argv){
int * q;
int * r;
euclidean(5,4, q, r);
printf("q = %d, r = %d", *q, *r);
return 0;
}
void euclidean(int a, int b, int * q, int * r)
{
*q = a / b;
*r = a % b;
//printf("q = %d, r = %d", *q, *r); //this won't show
return;
}
Both version aren't working. I compiled and ran the code on windows and the program is killed on runtime without printing anything (i.e. "q = 1, r = 4"). And my guess is if I had to compile and run it on linux, the terminal would gave me a "segmentation fault" error (not sure). I really don't see why it isn't working, especially with code listing 1. For code listing 2 one can argue that since the result of the operation are some sort of constant variable inside a function in which they were created, they could not be kept at the end of the function (I need confirmation on that too). Thanks

Answer 1

Keep It Simple, Stupid (there is such a principle).:)
#include<stdio.h>
#include<stdlib.h>
void euclidean(int a, int b, int * q, int * r);
int main(int argc, char ** argv){
int q;
int r;
euclidean(5,4, &q, &r);
printf("q = %d, r = %d", q, r);
return 0;
}
void euclidean(int a, int b, int * q, int * r)
{
*q = a / b;
*r = a % b;
}
If you want indeed to allocate memory in the function then the code will look like
#include<stdio.h>
#include<stdlib.h>
void euclidean(int a, int b, int * *q, int * *r);
int main(int argc, char ** argv){
int * q;
int * r;
euclidean(5,4, &q, &r);
printf("q = %d, r = %d", *q, *r);
free( q );
free( r );
return 0;
}
void euclidean(int a, int b, int * *q, int * *r){
*q = (int*)malloc(sizeof(int));
*r = (int*)malloc(sizeof(int));
**q = a/b;
**r = a % b;
}

Answer 2

You are not getting values for listing_2 because:-
void euclidean(int a, int b, int * q, int * r)
{
*q = a / b; <<<<<<<<<<<<<<<<<
*r = a % b;
//printf("q = %d, r = %d", *q, *r); //this won't show
return;
}
Neither you have allocated memory for q,r in main nor in euclidean. You were getting error as you are trying to deference a pointer not initialized to anything.

Answer 3

What you want is known as pass-by-reference, which allows the callee to modify the object the caller provides.
As C is purely pass-by-value, that is simulated with pointers:
The caller passes the address of the object the callee should modify, and the callee dereferences the received pointer to modify the intended target.
#include<stdio.h>
#include<stdlib.h>
void euclidean(int a, int b, int * q, int * r)
{
// Modify the target-objects q and r point to
*q = a / b;
*r = a % b;
}
int main() {
int q, r; // Not pointers, the objects themselves!
euclidean(5, 4, &q, &r); // Passing the addresses of q and r
printf("q = %d, r = %d", q, r);
}
No dynamic allocation at all.