How to return a pointer to an array in C?

Question

I'm writing a function which is to read in a file and insert the text character by character into an array, and then return a pointer to that array. The max size permitted for the file is supposted to be 2KB.

Here's what I have now:

int main(){
    char data[2048];
    char* data4=layer4(data);
}

char* layer4(char array[]){
    FILE *fp;
    fp=fopen("sendfile.txt","r+");
    fscanf(fp, "%c", array);
    for(int i=0; i<2048; i++){
        printf("%c\n",array[i]);
    }
    return &array;
}

Answer 1

You only need to return:

return array;

Answer 2

Multiple problems. Some implied by @Jay Kumar

char* layer4(char array[]){
    FILE *fp;
    fp=fopen("sendfile.txt","r+");

    // check if open succeeded
    if (fp == NULL) return NULL;   

    // Unclear why codes read only 1 char
    // fscanf(fp, "%c", array);
    // Suspect OP wants
    size_t i;
    for(i=0; i<2048; i++){
      if (fscanf(fp, "%c", &array[i]) != 1) break;
    }
    // Could have use fread()

    // More canonical to use type `size_t` than `int` to index arrays.
    // for(int i=0; i<2048; i++){
    for(size_t j=0; j<i; j++){
        printf("%c\n",array[j]);
    }

    // do not forget to put your toys away
    fclose(fp);

    // Simply return `array`
    // return &array;
    return array;
}

Answer 3

return a pointer to that array

No, you don't have to do that.

You can write your prototype as void layer4(char array[]) and modify the array directly - the change you made to array inside layer4 will retain after that function exits.

Answer 4

array is already pointer to char. So you need only to return array, not &array.

See this post Arrays decaying into pointers.

Answer 5

array is already a pointer. So just do: return array;

Answer 6

Your method should take the array as an array or as a pointer, as it is a pointer, then you can just return it as itself. try here

char* layer4(char array[])
{
    ..

    return array;
}