How to avoid memory leak where function returning an array

Question

In the following code pointer A had different address before calling the function copy(). Once the copy() is executed, pointer A got the address of pointer B which is declared in copy(). So, within main() when free(A) is executed, it's freeing the memory assigned to the pointer B in copy(). Now the problem is how to deallocate pointer A assigned within main() ? Is there any memory leaks in the following code? How to prevent it?

Here is the code:

#define size 10
int *copy(int *A){
int i;
int *B = (int *)calloc(size,sizeof(int));
printf("address of B=%p\n",B);
for(i=0;i<size;i++){
       B[i]=A[i]+1;
    }
for(i=0;i<size;i++){
   printf("%d ",B[i]);
   }
 printf("\n ");
 return B;
 }

 int main(){
     int i;
     int *A = (int *)calloc(size,sizeof(int));
     printf("address of A before copy()=%p\n",A);

     for(i=0;i<size;i++){
        A[i] = i;
        }
        A=copy(A);
      printf("address of A after copy()=%p\n",A);
     for(i=0;i<size;i++){
         printf("%d ",A[i]);    
         }
     printf("\n");
     free(A);
     return 0;
     }

And here is the output:

   address of A before copy()=0x1e64010
   address of B=0x1e64040
   1 2 3 4 5 6 7 8 9 10 
   address of A after copy()=0x1e64040
   1 2 3 4 5 6 7 8 9 10

tdao · Accepted Answer · 2016-02-14 05:49:44Z

4

Is there any memory leaks in the following code?

Yes, there's potentially a memory leak from these lines:

int *A = (int *)calloc(size,sizeof(int));

int *A = copy(A);   // <-- Do NOT allocate and then point the pointer to somewhere else

That is because you allocated memory for A, and then you pointed A to somewhere else, thus loosing the handle to free the memory initially allocated.

NOTE:

In this particular case, though, you might get away with it because it happened in main() - and the memory could be deallocated by the OS itself when the program exits.
Generally speaking, however, this kind of code would cause memory leak, especially when the function is not main(). Also, it's a bad practice leaving it the OS to do the clean up for you.

how to deallocate pointer A assigned within main() ?

How to make it works - here's a suggestion:

Declare a 2nd pointer B within main()
Then call copy to copy/modify content from original array.

The code should look like this:

 int *A = (int *)calloc(size,sizeof(int));

 // Initialize A
 for( i=0;i<size;i++ )
 {
    A[i] = i;
 }

 // Copy A to B, and modify the content.
 int *B = copy(A);   // <-- 2nd pointer

And at the end, free both pointers:

free( A );
free( B );

edited Feb 14, 2016 at 5:49

answered Feb 14, 2016 at 5:20

tdao

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2 Comments

precision Over a year ago

Thanks for your answer! I have one doubt here, what is the best way to update A with copy(A) before freeing B in main()?

tdao Over a year ago

If the only thing you wanted to is to modify A, then do not return int * in your copy function, simply use void copy(int *A) - that would allow you to modify A directly. Also change the name of the function to something else, not copy, maybe update() would be a better name.

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