Sum of two arrays using pointers in C

Question

I am trying to create a program in C that calculates the sum of two int arrays using pointers. Here is an example of what I want to do:

int a[] = {1,2,3,4}
int b[] = {1,2,3,4,5,6}
int c[] = sumArrays(a,b,4,6)
Output : c = {2,4,6,8,5,6}

The problem is my output is different, it shows:

Output : c = {2,4,6,8}

Any idea what did I do wrong and how to correct it ? Here is my code :

#include <stdio.h>
#include <stdlib.h>

int* sumArrays(int *arr1, int *arr2, int dim1,int dim2)
{
    int *ret = NULL;

    if(dim1<dim2) ret = (int*) malloc(dim2*sizeof(int));
    else ret =  (int*) malloc(dim1*sizeof(int));
    if(ret)
    {
        if(dim1<dim2) {
            int i = 0;
            for (i = 0;i < dim1;i++) {
                ret[i] = arr1[i] + arr2[i];
            }
            for (i = dim1; i < dim2;i++) {
                ret[i]=arr2[i];
            }   
        } else {
            int i = 0;
            for (i = 0;i < dim2;i++) {
                ret[i] = arr1[i] + arr2[i];
            }
            for (i = dim2; i < dim1;i++) {
                ret[i]=arr1[i];
            } 
        }
    }
    return ret;
}


int main()
{
int *a[] = {1,2,3,4};
int *b[] = {1,2,3,4,5,6};

int *c = sumArrays(a,b,4,6);
printf("c = ");
int i;
for (i = 0; i < sizeof(c); i++) {
    printf("%d,",c[i]);
} 
}

On another note, you should be getting c = {2,4,6,8,}, not c = {2,4,6,8} — anishsane
– anishsane, Commented May 14, 2017 at 16:45
Btw, the code is incorrect. It could lead to undefined behaviour. You are allocating some memory inside a function and when you go out of its scope, who knows. Never do that — mik1904
– mik1904, Commented May 14, 2017 at 17:04

Andriy Berestovskyy · Accepted Answer · 2017-05-14 16:20:29Z

5

Sizeof c will always return 4 for 32-bit system and 8 for 64-bit, because c is a pointer to int.

So to print the result array you should write:

for (i = 0; i < 6; i++) {

answered May 14, 2017 at 16:20

Andriy Berestovskyy

8,6243 gold badges21 silver badges35 bronze badges

Sign up to request clarification or add additional context in comments.

1 Comment

nicoolaslb Over a year ago

That is so obvious.. Thank you !

Soner from The Ottoman Empire · Accepted Answer · 2017-05-14 16:55:17Z

You have problem about using pointers.

int *a[] = {1,2,3,4};      // a is an array of pointers to integers.
int *b[] = {1,2,3,4,5,6};  // b is an array of pointers to integers.

By doing this, you are declaring array of pointers to integers. So, it most likely causes compiler warning or error depending on your compile settings like, initialization makes pointer from integer without a cast. When you are passing the actual arguments to formal parameters, it causes same warning as well. Your main should be like this,

int main()
{
    int a[] = {1,2,3,4}; 
    int b[] = {1,2,3,4,5,6};
    int sizeA = sizeof(a) / sizeof(*a);
    int sizeB = sizeof(b) / sizeof(*b);


    int *c = sumArrays(a,b,sizeA,sizeB);
    printf("c = ");
    int i;
    for (i = 0; i < (sizeA < sizeB ? sizeB : sizeA); i++) {
        printf("%d,",c[i]);
    }
}

Collectives™ on Stack Overflow

Sum of two arrays using pointers in C

2 Answers 2

1 Comment

Comments

Your Answer

Hot Network Questions

Collectives™ on Stack Overflow

2 Answers 2

1 Comment

Comments

Your Answer

Sign up or log in

Post as a guest

Related