Segmentation fault while returning pointer

Question

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.

Dereferencing a null pointer is not legal.

Some programmer dude
– Some programmer dude

2016-05-17 06:34:17 +00:00
Commented May 17, 2016 at 6:34 — Some programmer dude
– Some programmer dude, Commented May 17, 2016 at 6:34

LPs · Accepted Answer · 2016-05-17 07:14:10Z

5

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;
}

edited May 17, 2016 at 7:14

answered May 17, 2016 at 6:36

LPs

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4 Comments

Jonathan Leffler Over a year ago

And there's really no benefit to passing the arguments as pointers, so you could use static inline int sum(int p, int q) { return p + q; } if you wanted to, though you'd probably want to place that above where it is used.

LPs Over a year ago

@JonathanLeffler Yes, sure, but OP wrote: I have begin learning pointers in C

Jonathan Leffler Over a year ago

And one of the lessons to learn about pointers is when to use them and when not to use them.

Ozitrick Over a year ago

Thanks! This is really helpful. Using malloc solves this issue. I did came up with alternatives, however I wanted to implement function returning a pointer specifically. Thanks for your help!

Haris · Accepted Answer · 2016-05-17 08:13:43Z

4

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;
}

edited May 17, 2016 at 8:13

answered May 17, 2016 at 6:31

Haris

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3 Comments

Jonathan Leffler Over a year ago

Why not: int *result = malloc(sizeof(*result));? And it might be an idea to use if (result != NULL) *result = *p + *q;. There's also no good reason to pass the arguments by pointer rather than just by value.

Haris Over a year ago

@JonathanLeffler. Thanks. Added the return value check.

Haris Over a year ago

@M.M, All set now. :)

Some programmer dude · Accepted Answer · 2016-05-17 07:01:01Z

2

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.

answered May 17, 2016 at 7:01

Some programmer dude

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1 Comment

LPs Over a year ago

I have taken the liberty to add code to my answer based on your answer.

BikyMandal · Accepted Answer · 2021-04-24 22:46:32Z

1

#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

answered Apr 24, 2021 at 22:46

BikyMandal

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