Copy array to dynamically allocated memory

Question

I got my code working, but i feel as if there is a faster way to do this especially in my function copy. here is my code. can this be any faster? this is in C btw. Also when i return cpy from the function does it delete the dynamic memory since its out of scope? I don't want to have memory leaks :P

#include <stdio.h>
#include <stdlib.h>
double *copy(double a[], unsigned ele);
int main(){
    double arr[8], *ptr;
    unsigned i=0;
    for(;i<7;i++){
        scanf_s("%lf", &arr[i]);
    }
    ptr=copy(arr, 8);
    for(i=0;i<7; i++)
        printf("%f", ptr[i]);

}

double *copy(double a[], unsigned ele){
    double *cpy= malloc(sizeof(double)*ele);
    int i=0;
    for(;i<ele; i++)
        cpy[i]=a[i];
    return cpy;
}

Solution 1

You can replace your for with a memcpy.

memcpy(cpy, a, ele * sizeof *cpy);

Other than that what you're doing it pretty okay: you're returning a pointer, thus the caller has a chance to free it.

Also I consider it good practice NOT to call free if you're going to exit - the OS will collect the memory anyway.

Solution 2

Use Memcpy. It might be able to take advantage of your underlying hardware etc... No need to reinvent the wheel. :-)

void * memcpy ( void * destination, const void * source, size_t num );


double *copy(double a[], unsigned ele){
    size_t size = sizeof(double)*ele ;
    double *cpy= malloc(size);
    memcpy( cpy, a, size ) ;
    return cpy;
}

Solution 3

does it delete the dynamic memory since its out of scope

No, it doesn't: the only memory that goes out of scope is the one holding the pointer; you return the value of that pointer, and it's the only thing you need to access the memory you malloc-ed and which stays malloc-ed until you free it explicitly.

I don't want to have memory leaks

Memory leaks happen when you get memory that you won't release. Indeed, your code have a memory leak: you do not free the memory the pointer is returned by copy: free(ptr); at the end of your code. In this case, it does not matter too much, but it's a good practice to avoid to forget it.

Note also that if the memory would be freed automatically when a pointer to it goes out of scope, it would be freed from inside the copy function itself and you would return an invalid pointer; and then there would be a lot of double free and alike, each time a pointer goes out of scope! Fortunately this does not happen since memory you got using malloc must be explicitly freed with free.