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 (let’s 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.

Proposed Code Format for Array Returning Function

char *returnArray(char array []){
  char returned [10];
  // Methods to pull values from the array, interpret
  // them, and then create a 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

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 2

C's treatment of arrays is very different from Java's, and you'll have to adjust your thinking accordingly. Arrays in C are not first class objects (that is, an array expression does not retain its "array-ness" in most contexts). In C, an expression of type "N-element array of T" will be implicitly converted ("decay") to an expression of type "pointer to T", except when the array expression is an operand of the sizeof or unary & operators, or if the array expression is a string literal being used to initialize another array in a declaration.

Among other things, this means that you cannot pass an array expression to a function and have it received as an array type; the function actually receives a pointer type:

void foo(char *a, size_t asize)
{
  // do something with a
}

int bar(void)
{
  char str[6] = "Hello";
  foo(str, sizeof str);
}

In the call to foo, the expression str is converted from type char [6] to char *, which is why the first parameter of foo is declared char *a instead of char a[6]. In sizeof str, since the array expression is an operand of the sizeof operator, it's not converted to a pointer type, so you get the number of bytes in the array (6).

If you're really interested, you can read Dennis Ritchie's The Development of the C Language to understand where this treatment comes from.

The upshot is that functions cannot return array types, which is fine since array expressions cannot be the target of an assignment, either.

The safest method is for the caller to define the array, and pass its address and size to the function that's supposed to write to it:

void returnArray(const char *srcArray, size_t srcSize, char *dstArray, char dstSize)
{
  ...
  dstArray[i] = some_value_derived_from(srcArray[i]);
  ...
}

int main(void)
{
  char src[] = "This is a test";
  char dst[sizeof src];
  ...
  returnArray(src, sizeof src, dst, sizeof dst);
  ...
}

Another method is for the function to allocate the array dynamically and return the pointer and size:

char *returnArray(const char *srcArray, size_t srcSize, size_t *dstSize)
{
  char *dstArray = malloc(srcSize);
  if (dstArray)
  {
    *dstSize = srcSize;
    ...
  }
  return dstArray;
}

int main(void)
{
  char src[] = "This is a test";
  char *dst;
  size_t dstSize;

  dst = returnArray(src, sizeof src, &dstSize);
  ...
  free(dst);
  ...
}

In this case, the caller is responsible for deallocating the array with the free library function.

Note that dst in the above code is a simple pointer to char, not a pointer to an array of char. C's pointer and array semantics are such that you can apply the subscript operator [] to either an expression of array type or pointer type; both src[i] and dst[i] will access the i'th element of the array (even though only src has array type).

You can declare a pointer to an N-element array of T and do something similar:

char (*returnArray(const char *srcArr, size_t srcSize))[SOME_SIZE]
{
  char (*dstArr)[SOME_SIZE] = malloc(sizeof *dstArr);
  if (dstArr)
  {
    ...
    (*dstArr)[i] = ...;
    ...
  }
  return dstArr;
}

int main(void)
{
  char src[] = "This is a test";
  char (*dst)[SOME_SIZE];
  ...
  dst = returnArray(src, sizeof src);
  ...
  printf("%c", (*dst)[j]);
  ...
}

There are several drawbacks with the above. First of all, older versions of C expect SOME_SIZE to be a compile-time constant, meaning that function will only ever work with one array size. Secondly, you have to dereference the pointer before applying the subscript, which clutters the code. Pointers to arrays work better when you're dealing with multi-dimensional arrays.

Answer 3

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;
    }

Answer 4

Use 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 5

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 every time 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 then 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 6

In your case, you are creating an array on the stack and once you leave the function scope, the array will be deallocated. Instead, create a dynamically allocated array and return a pointer to it.

char * returnArray(char *arr, int size) {
    char *new_arr = malloc(sizeof(char) * size);
    for(int i = 0; i < size; ++i) {
        new_arr[i] = arr[i];
    }
    return new_arr;
}

int main() {

    char arr[7]= {1,0,0,0,0,1,1};
    char *new_arr = returnArray(arr, 7);

    // don't forget to free the memory after you're done with the array
    free(new_arr);

}

Answer 7

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 8

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.