free() in dynamic memory - c

Question

Do I use correctly free() in the code below? Is it a memory leak? Is it a problem use free() in the main part and not in the function? If yes there is a method to free in the function and not in main?
This code copy an array in another one.
int *copy(const int *arr,int n);
int main(){
int *p_arr1,*p_arr2;
int n,i;
printf("Insert size of array: ");
scanf("%d",&n);
p_arr1 = calloc(n,sizeof(int));
for(i=0;i<n;i++){
printf("Insert element %d of the array: ",i+1);
scanf("%d",p_arr1+i);
}
p_arr2 = copy(p_arr1,n);
for(i=0;i<n;i++){
printf("%d ",*p_arr2);
p_arr2++;
}
free(p_arr1);
free(p_arr2);
return 0;
}
int *copy(const int *arr,int n){
int i;
int *new;
new = calloc(n, sizeof(int));
for(i=0;i<n;i++){
new[i] += arr[i];
}
return new;
}

Answer 1

As long as you have the pointer returned by malloc (or in your case calloc) you can pass it to free when and wherever you want, it doesn't have to be in the same function.
However, after the loop where you print the contents of p_arr2, you no longer have the pointer returned by calloc inside the function, because you modify the pointer in the loop.
You need to use a temporary pointer variable for the loop:
int *p_arr2_tmp = p_arr2;
for (size_t i = 0; i < n; ++i)
{
printf("%d ", *p_arr2_tmp);
++p_arr2_tmp;
}
// Now we can free the memory pointed to by the original p_arr2 pointer
free(p_arr2);
Or you could use simple array indexing instead:
for (size_t i = 0; i < n; ++i)
{
printf("%d ", p_arr2[i]);
}
// The pointer p_arr2 wasn't modified, so it can be passed to free
free(p_arr2);

Answer 2

In this loop
for(i=0;i<n;i++){
printf("%d ",*p_arr2);
p_arr2++;
}
the value of the pointer p_arr2 is being changed, So using the changed pointer in a call of free results in undefined behavior.
You should write
for(i=0;i<n;i++){
printf("%d ", p_arr2[i] );
}
Also it is unclear why you are using the compound operator += in the function instead of the operator =.
new[i] += arr[i];
The function can be defined the following way
int * copy( const int *arr, size_t n )
{
int *new_arr = malloc( n * sizeof( int ) );
if ( new_arr != NULL )
{
memcpy( new_arr, arr, n * sizeof( int ) );
}
return new_arr;
}
If you want to use a pointer in the loop that outputs the newly created array then it can look the following way
for ( const int *p = p_arr2; p != p_arr2 + n; ++p )
{
printf( "%d ",*p );
}
putchar( '\n' );
If the aim is to write a program that uses only pointers and excludes using of the subscript operator and indices then your program can look the following way
#include <stdio.h>
#include <stdlib.h>
int * copy( const int *arr, size_t n )
{
int *new_arr = malloc( n * sizeof( int ) );
if ( new_arr != NULL )
{
for ( int *p = new_arr; p != new_arr + n; ++p )
{
*p = *arr++;
}
}
return new_arr;
}
int main(void)
{
size_t n;
printf( "Insert size of array: " );
scanf( "%zu", &n );
int *p_arr1 = calloc( n, sizeof( int ) );
for ( int *p = p_arr1; p != p_arr1 + n; ++p )
{
printf( "Insert element %d of the array: ", ( int )( p - p_arr1 + 1 ) );
scanf( "%d", p );
}
int *p_arr2 = copy( p_arr1, n );
if ( p_arr2 != NULL )
{
for ( const int *p = p_arr2; p != p_arr2 + n; ++p )
{
printf( "%d ",*p );
}
putchar( '\n' );
}
free( p_arr2 );
free( p_arr1 );
return 0;
}
The program output might look like
Insert size of array: 10
Insert element 1 of the array: 0
Insert element 2 of the array: 1
Insert element 3 of the array: 2
Insert element 4 of the array: 3
Insert element 5 of the array: 4
Insert element 6 of the array: 5
Insert element 7 of the array: 6
Insert element 8 of the array: 7
Insert element 9 of the array: 8
Insert element 10 of the array: 9
0 1 2 3 4 5 6 7 8 9