CODEWITHSUNDEEP
cpointerssegmentation-fault
Ozitrick
Ozitrick

Reputation: 87

Segmentation fault while returning pointer

I have begin learning pointers in C.

When I try to return pointer in a function, I'm getting segmentation fault error.

Here is the code : 

#include<stdio.h>

int *sum(int *, int *);

int main(void)
{
    int a, b;
    int *ans = NULL;
    printf("Enter number a : ");
    scanf("%d", &a);
    printf("Enter number b : ");
    scanf("%d", &b);

    ans = sum(&a, &b);
    printf("Sum = %d", *ans); 

    return 0;   
}

int *sum(int *p, int *q)
{
    int *result = NULL;
    *result = *p + *q;
    return (result);
}

And the output : 

Enter number a : 10
Enter number b : 20
Segmentation fault

Segmentation fault occurs in sum function, when result is declared as pointer. However, I am unable to figure out the reason for the same. Any help regarding this is really appreciable.

Upvotes: 1

Views: 2567

Answers (4)

BikyMandal
BikyMandal

Reputation: 119

#include<stdio.h>

int *sum(int *, int *);

int main(void)
{
    int a, b;
    printf("Enter number a : ");
    scanf("%d", &a);
    printf("Enter number b : ");
    scanf("%d", &b);

    int* ans = sum(&a, &b);
    printf("Sum = %d", *ans); 

    return 0;   
}

int *sum(int *p, int *q)
{
    int plus = *p + *q;
    int *ans = &plus;
    return ans;
}
Store the Sum in a Different variable plus and create a pointer *ans that points to that variable plus and return the pointer variable ans that holds the address of plus

Upvotes: 1

Haris
Haris

Reputation: 12272

Allocate memory before trying to store anything, and check the return of malloc()

int *result = NULL;
result = malloc(sizeof(*result));
if(result != NULL)
    *result = *p + *q;
else
    printf("malloc returned error");

Also, check the return of the function in main() and exit accordingly.

int main(void)
{
    .
    .
    .
    ans = sum(&a, &b);
    if(ans == NULL)
        return 0;

    printf("Sum = %d\n", ans);
    free(ans);    //free the memory then
    return 0;
}

Upvotes: 4

LPs
LPs

Reputation: 16243

You are initing a pointer to NULL and then you are deferencing it: it is Undefined behavior

Change sum function to

int *sum(int *p, int *q)
{
    int *result = malloc(sizeof(int));

    // check if malloc returned a valid pointer before to dereference it
    if (result != NULL)
    {
        *result = *p + *q;
    }

    return (result);
}

and add a free call to free the allocated memory.

    // check if sum function allocate the pointer before to dereference it
    if (ans != NULL)
    {
       printf("Sum = %d", *ans); 
    }

    free(ans);

    return 0;   
}

You could also avoid to use pointer to return value:

#include<stdio.h>

int sum(int *, int *);

int main(void)
{
    int a, b;
    int ans;
    printf("Enter number a : ");
    scanf("%d", &a);
    printf("Enter number b : ");
    scanf("%d", &b);

    ans = sum(&a, &b);
    printf("Sum = %d\n", ans); 

    return 0;   
}

int sum(int *p, int *q)
{
    int result = *p + *q;
    return (result);
}

The sum function could also be like:

int sum (int *p, int *q)
{
    return (*p + *q);
}

EDIT

As @JonathanLeffler wrote in his answer you can do also:

#include<stdio.h>

void sum(int *, int *, int *);

int main(void)
{
    int a, b;
    int ans;
    printf("Enter number a : ");
    scanf("%d", &a);
    printf("Enter number b : ");
    scanf("%d", &b);

    sum(&ans, &a, &b);
    printf("Sum = %d\n", ans);

    return 0;
}

void sum(int *result, int *p, int *q)
{
    *result = *p + *q;
}

Upvotes: 5

Some programmer dude
Some programmer dude

Reputation: 409432

A third alternative is to declare ans as a normal int variable in the main function and pass a pointer to it to the sum function, like you do with the other two arguments. This is actually emulating call by reference.

Upvotes: 2

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