Segmentation fault in return index

Question

int* twoSum(int* nums, int numsSize, int target, int* returnSize){

    int i = 0, j = 0;

    for(i=0; i < numsSize; i++)
    {
        for(j = i+1; j < numsSize; j++)
        {
            if(nums[j] == target - nums[i])
            {
                int *index = (int *)malloc(sizeof(int) * 2);
                index[0] = i;
                index[1] = j;
                return index;
            }
        }
    }
    return NULL;
}

I was writing the code for the following

Given nums = [2, 7, 11, 15], target = 9,

Because nums[0] + nums[1] = 2 + 7 = 9, return [0, 1].

I tried debugging the code, i and j values 0 and 1 as expected. But now I'am facing the issue while returning the index. It is showing as segmentation fault. Can anyone please correct the above code?

Run your program with `gdb` and then run `backtrace`. It will show you which line it is segfaulting on. — Gillespie, Jan 28 '20 at 18:31
The `return index` instruction cannot possibly segfault. Your code looks right, there's no problem with it. You need to add a [mcve] to your question. Your problem is somewhere else. — Marco Bonelli, Jan 28 '20 at 18:38
your code is OK .. you have problem somewhere else. [my solution](https://onlinegdb.com/S1qJXbA-L) — houssam, Jan 28 '20 at 18:53
I´ve seen you use `NULL` as return value: https://stackoverflow.com/questions/59437282/can-i-use-null-as-substitution-for-the-value-of-0 — RobertS supports Monica Cellio, Jan 28 '20 at 19:41
@houssam I just saw your solution. But I didn't get what is purpose of typecasting malloc to void. Can you please explain? — chaya kumar, Jan 28 '20 at 19:44
@chayakumar : many compilers throw warning `Warning: incompatible implicit declaration of built-in function ‘malloc` if you use `malloc` without including the header file: `#include `. but the code works !!. You can use it without casting such as `int *index = malloc(sizeof(int) * 2);`. see : [Do I cast the result of malloc?](https://stackoverflow.com/q/605845/992406) — houssam, Jan 28 '20 at 20:07

score 1 · Answer 1 · answered Jan 28 '20 at 18:41

1

Looks like codeforces or similar platform problems. Segmentation Fault is happening because you have not set returnSize passed by caller as same is required for freeing returned array. Thus solution would be to set returnSize to 2.

answered Jan 28 '20 at 18:41

H S

1,211
6
11

but `returnSize` still not being used inside the function. I'd consider to remove though. – Muzol Jan 28 '20 at 19:02
@Muzol I agree. However, this code will be judged by online judge which requires `returnSize` as a param. That's just how it's supposed to work. – H S Jan 28 '20 at 19:05

score 0 · Answer 2 · answered Jan 28 '20 at 20:31

#include <stdio.h>
#include <stdlib.h>
int* twoSum(int* nums, int numsSize, int target, int* returnSize)
{
    int i = 0, j = 0;
    for(i=0; i < numsSize; i++)
    {
        for(j = i+1; j < numsSize; j++)
        {
            printf("i=%d j=%d \n",i,j);
            if(nums[j] == target - nums[i])
            {
                int *index = malloc(sizeof(int) * 2);
                index[0] = i;
                index[1] = j;
                return index;
            }
        }
    }
    return NULL;
}

int main()
{
    int nums[] = {2, 7, 11, 15}; // works
    //int * nums = {2, 7, 11, 15}; //compile OK (some compilers) but Segmentation fault .
    int target = 9;
    int * result = twoSum(nums,4,target,NULL);
    if(result)
        printf("result[0]=%d result[1]=%d" ,result[0] , result[1]);
    return 0;
}

Your code is correct; but it depends on how you define nums; is it array or pointer?
Initializing array (pointer) with values
Is an array name a pointer?

Segmentation fault in return index

2 Answers2