The exercise says "Make a function with parameters two int arrays and k which is their size. The function should return another array (size k) where every element of it is the sum of the two arrays of the same position. That's what I wrote, but it crashes. Do I have to do it with pointers?

``#include <stdio.h>#include <stdlib.h>void sumarray(int k,int A[k],int B[k]){int sum[k],i;for(i=0;i<k;i++){sum[i]=A[i]+B[i];printf("sum[%d]=%d\n",i,sum[i]);}}main(){int i,g,a[g],b[g];printf("Give size of both arrays: ");scanf("%d",&g);for(i=0;i<g;i++){a[i]=rand();b[i]=rand();}sumarray(g,a,b);system("pause");}``

Example: If i have A={1,2,3,4} and B={4,3,2,1} the program will return C={5,5,5,5).

This:

``````int i,g,a[g],b[g];
``````

causes undefined behaviour. The value of `g` is undefined upon initialisation, so therefore the length of `a` and `b` will be undefined.

You probably want something like:

``````int i, g;
int *a;
int *b;  // Note: recommend declaring on separate lines, to avoid issues
scanf("%d", &g);
a = malloc(sizeof(*a) * g);
b = malloc(sizeof(*b) * g);
...
free(a);
free(b);
``````

Its impossible to first do `a[g]` when dynamically getting `g`.

Your first lines in main should be:

``````int i,g;
int *a,*b;
printf("Give size of both arrays: ");
scanf("%d",&g);
a = (int *)malloc(g*sizeof(int));
b = (int *)malloc(g*sizeof(int));
``````
``````int sum[k] ;
``````

`k` is a variable but the size of the array should be a constant.

The function should return another array (size k) ...

But the function you wrote returns `void` which is clearly wrong.

Do I have to do it with pointers?

Yes.

One issue is that you've attempted to declare dynamically sized arrays on the stack (e.g. `a[g]`). You need to declare pointers for each array and then dynamically allocate your variable sized array once you've read in the value of `g`.

change the function summary signature (the definition part of the declaration) to this and try it out:

`void sumarray(int k,int* A,int* B){`

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