Returning an array using C

Question

I am relatively new to C and I need some help with methods dealing with arrays. Coming from Java programming, I am used to being able to say int [] method()in order to return an array. However, I have found out that with C you have to use pointers for arrays when you return them. Being a new programmer, I really do not understand this at all, even with the many forums I have looked through.

Basically, I am trying to write a method that returns a char array in C. I will provide the method (lets call it returnArray) with an array. It will create a new array from the previous array and return a pointer to it. I just need some help on how to get this started and how to read the pointer once it is sent out of the array. Any help explaining this is appreciated.

Proposed Code Format for Array Returning Function

char *returnArray(char array []){
 char returned [10];
 //methods to pull values from array, interpret them, and then create new array
 return &(returned[0]); //is this correct?
} 

Caller of the Function

int main(){
 int i=0;
 char array []={1,0,0,0,0,1,1};
 char arrayCount=0;
 char* returnedArray = returnArray(&arrayCount); ///is this correct?
 for (i=0; i<10;i++)
  printf(%d, ",", returnedArray[i]);  //is this correctly formatted?
}

I have not tested this yet as my C compiler is not working at the moment but I would like to figure this out

Answer 1

How about this deliciously evil implementation?

array.h

#define IMPORT_ARRAY(TYPE)    \
    \
struct TYPE##Array {    \
    TYPE* contents;    \
    size_t size;    \
};    \
    \
struct TYPE##Array new_##TYPE##Array() {    \
    struct TYPE##Array a;    \
    a.contents = NULL;    \
    a.size = 0;    \
    return a;    \
}    \
    \
void array_add(struct TYPE##Array* o, TYPE value) {    \
    TYPE* a = malloc((o->size + 1) * sizeof(TYPE));    \
    TYPE i;    \
    for(i = 0; i < o->size; ++i) {    \
        a[i] = o->contents[i];    \
    }    \
    ++(o->size);    \
    a[o->size - 1] = value;    \
    free(o->contents);    \
    o->contents = a;    \
}    \
void array_destroy(struct TYPE##Array* o) {    \
    free(o->contents);    \
}    \
TYPE* array_begin(struct TYPE##Array* o) {    \
    return o->contents;    \
}    \
TYPE* array_end(struct TYPE##Array* o) {    \
    return o->contents + o->size;    \
}

main.c

#include <stdlib.h>
#include "array.h"

IMPORT_ARRAY(int);

struct intArray return_an_array() {
    struct intArray a;
    a = new_intArray();
    array_add(&a, 1);
    array_add(&a, 2);
    array_add(&a, 3);
    return a;
}

int main() {
    struct intArray a;
    int* it;
    int* begin;
    int* end;
    a = return_an_array();
    begin = array_begin(&a);
    end = array_end(&a);
    for(it = begin; it != end; ++it) {
        printf("%d ", *it);
    }
    array_destroy(&a);
    getchar();
    return 0;
}

Answer 2

I am not saying that this is the best solution or a preferred solution to the given problem. However, it may be useful to remember that functions can return structs. Although functions cannot return arrays, arrays can be wrapped in structs and the function can return the struct thereby carrying the array with it. This works for fixed length arrays.

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

    typedef
    struct 
    {
        char v[10];
    } CHAR_ARRAY;



    CHAR_ARRAY returnArray(CHAR_ARRAY array_in, int size)
    {
        CHAR_ARRAY returned;

        /*
        . . . methods to pull values from array, interpret them, and then create new array
        */

        for (int i = 0;  i < size; i++ )
            returned.v[i] = array_in.v[i] + 1;

        return returned; // Works!
    } 




    int main(int argc, char * argv[])
    {
        CHAR_ARRAY array = {1,0,0,0,0,1,1};

        char arrayCount = 7;

        CHAR_ARRAY returnedArray = returnArray(array, arrayCount); 

        for (int i = 0; i < arrayCount; i++)
            printf("%d, ", returnedArray.v[i]);  //is this correctly formatted?

        getchar();
        return 0;
    }

I invite comments on the strengths and weaknesses of this technique. I have not bothered to do so.

Answer 3

You can't return arrays from functions in C. You also can't (shouldn't) do this:

char *returnArray(char array []){
 char returned [10];
 //methods to pull values from array, interpret them, and then create new array
 return &(returned[0]); //is this correct?
} 

returned is created with automatic storage duration and references to it will become invalid once it leaves its declaring scope, i.e., when the function returns.

You will need to dynamically allocate the memory inside of the function or fill a preallocated buffer provided by the caller.

Option 1:

dynamically allocate the memory inside of the function (caller responsible for deallocating ret)

char *foo(int count) {
    char *ret = malloc(count);
    if(!ret)
        return NULL;

    for(int i = 0; i < count; ++i) 
        ret[i] = i;

    return ret;
}

Call it like so:

int main() {
    char *p = foo(10);
    if(p) {
        // do stuff with p
        free(p);
    }

    return 0;
}

Option 2:

fill a preallocated buffer provided by the caller (caller allocates buf and passes to the function)

void foo(char *buf, int count) {
    for(int i = 0; i < count; ++i)
        buf[i] = i;
}

And call it like so:

int main() {
    char arr[10] = {0};
    foo(arr, 10);
    // No need to deallocate because we allocated 
    // arr with automatic storage duration.
    // If we had dynamically allocated it
    // (i.e. malloc or some variant) then we 
    // would need to call free(arr)
}

Answer 4

You can do it using heap memory (through malloc() invocation) like other answers reported here, but you must always manage the memory (use free() function everytime you call your function). You can also do it with a static array:

char* returnArrayPointer() 
{
static char array[SIZE];

// do something in your array here

return array; 
}

You can than use it without worrying about memory management.

int main() 
{
char* myArray = returnArrayPointer();
/* use your array here */
/* don't worry to free memory here */
}

In this example you must use static keyword in array definition to set to application-long the array lifetime, so it will not destroyed after return statement. Of course, in this way you occupy SIZE bytes in your memory for the entire application life, so size it properly!

Answer 5

Your method will return a local stack variable that will fail badly. To return an array, create one outside the function, pass it by address into the function, then modify it, or create an array on the heap and return that variable. Both will work, but the first doesn't require any dynamic memory allocation to get it working correctly.

void returnArray(int size, char *retArray)
{
  // work directly with retArray or memcpy into it from elsewhere like
  // memcpy(retArray, localArray, size); 
}

#define ARRAY_SIZE 20

int main(void)
{
  char foo[ARRAY_SIZE];
  returnArray(ARRAY_SIZE, foo);
}

Answer 6

You can use code like this:

char *MyFunction(some arguments...)
{
    char *pointer = malloc(size for the new array);
    if (!pointer)
        An error occurred, abort or do something about the error.
    return pointer; // Return address of memory to the caller.
}

When you do this, the memory should later be freed, by passing the address to free.

There are other options. A routine might return a pointer to an array (or portion of an array) that is part of some existing structure. The caller might pass an array, and the routine merely writes into the array, rather than allocating space for a new array.