Returning Arrays/Pointers from a function

Question

I am trying to create a new integer array which is derived from a string of characters. For example :

char x[] = \"12334 23845 32084\";  

int y[] = { 12334,         


        

Answer 1

int * splitString( char string[], int n )
{
    int newArray[n];
    return ( newArray );
}

This is very bad! The array newArray local to the function gets destroyed when the function returns. You'd be left out with a dangling pointer and using it would invoke undefined behaviour.

You can't return an array from a function. The best you can do is

int * splitString( char string[], int n )
{
    int *newArray = malloc(n*sizeof(int)); // the array gets allocated on the heap rather than on the stack(1)
    // Code 
    return ( newArray );
}

Don't forget to free the allocated memory.

(1) Note that the standard doesn't use/define the term stack or heap as such.

Answer 2

The main issue i see is trying to return memory which you allocated on the stack, which becomes invalid once the function it was allocated in reutrns (in this case your splitString). What you can do is allocate the memory in the caller, and pass a pointer to the beginning of the array into the function

/* Convert string of integers into int array. */
void splitString(char string[], int *out_arr, int n )
{

    // code that fills each cell of out_arr

}

int main( void )
{
    int x[n];
    splitString( string,(int *)x, n );

    return ( 0 );
}

Answer 3

Rather than returning an array with return (newArray), you return a pointer to the first element of newArray.

The problem is that you're allocating the array the wrong way. If you instantiate it with int newArray[n], memory gets allocated on the current stack frame. That memory will be freed as soon as your function returns, and whatever was in the array will be garbage. Instead, do the following:

int *newArray = malloc(n * sizeof(int));
// etc.
return newArray

By using malloc, you allocate memory on the heap, where it will survive past the end of the current stack frame. Just remember to free(newArray) somewhere in your program when you're done.

Answer 4

You can wrap an array in a structure and then return an instance of the structure. I'm mentioning this for completeness, it's not really something you'd want to do as it's ugly and there are better alternatives.

#include <stdio.h>

struct retval
{
    int a[10];
};

struct retval test()
{
    struct retval v = {{1, 5, 6}};
    return v;
}

int main()
{
    struct retval data = test();
    printf("%d %d\n", data.a[1], data.a[2]);
}

Answer 5

Typically, you require the caller to pass in the result array.

void splitString( const char string[], int result[], int n) {
    //....
}

This is advantageous because the caller can allocate that memory wherever they want.

Answer 6

The problem is you're returning a pointer to something on the stack. You need to create your array on the heap, then free it when you're done:

int * splitString( char string[], int n )
{
    int *newArray = malloc(sizeof(int) * n);

    // CODE

    return ( newArray );
}

int main( void )
{
    int *x = splitString( string, n );

    // use it

    free(x);

    return ( 0 );
}