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

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)
}