Why will this not print? - c

Question

Before you feel the need to mark this as a duplicate post, please don't. I have read all the threads on pointers, arrays, and functions I could find but almost all of them are far too advanced to be of any help to me.
I'm not getting an error, however my code will not print my array. It seems the issue here is using scanf. I don't think the values entered are actually being put into the array in main(). I've tried using pointers, but then I get the error "Thread 1: EXC_BAD_ACCESS (code=1, address=0x0)" whenever I try to use scanf to collect user inputted values to put into the array.
What I am working on is limited to declaring my array in the main() function, but all the operations are to be performed in promptData() function. Any help would be great, I'm at my wits end trying to figure this out on my own.
#import <stdio.h>
void promptData(double data[], int numElem);
int main(int argc, const char * argv[])
{
int size, i;
double array[size];
promptData(array, size);
for (i = 0; i < size; i++)
printf("%.2lf\n", array[i]);
return 0;
}
void promptData(double data[], int numElem)
{
int i;
printf("Enter integer values for size of array.\n");
scanf("%i", &numElem);
for (i = 0; i < numElem; i++)
{
printf("Enter array values.\n");
scanf("%lf", &data[i]);
}
}

Answer 1

Your program has undefined behaviour because variable size was not initialized and has indeterminate value.
You should at first in main ask the user to enter the size of the array then define the array itself and only after that fill it with values.
For example
int main(int argc, const char * argv[])
{
int size = 0;
printf( "Enter a positive integer value for the size of the array: ");
scanf( "%i", &size);
if ( size == 0 ) exit( 1 );
double array[size];
promptData(array, size);
//...
Also in C there is no such a directive as
#import <stdio.h>
Use instead
#include <stdio.h>

Answer 2

At least in ANSI C 89 and C 90, you can't give a variable as the size of an array. The size of array should be known at compile time. You should be doing something like double array[size];.
Even in C99, where you can have variable sized arrays; the variables should contain proper index values at the time you declare the array. In that case, you should read the number from stdin and then declare the array.
Also in C, all parameters are passed by value. This means every function takes a copy of the parameters in the function. If you want to modify a variable's value, you should pass a pointer to it, and then modify the pointer's dereferenced value, something like:
void change(int *x)
{
*x = 7;
}
void first(void)
{
int x = 5;
change(&x);
printf("%d\n", x);
}

Answer 3

Adding on to the other, correct, answer by Zenith, if you want a dynamically allocated array (like you want to be able to change its size based on user input), then your only option is to use one of the memory allocation functions like malloc().
Once you actually have the size in your main function, declare your array like this:
int *myArray = malloc(sizeof(int) * size));//note that malloc will return a NULL if it fails
//you should always check
if(myArray != null) {
//do stuff with myArray like you were. You can just use myArray[] as long as you
//make SURE that you don't go beyond 'size'
}
free(myArray);
//VERY important that every malloc() has a free() with it
Note: untested, but the idea is there.
Further, to answer your other question.
If you find yourself in a situation where you need to call a function and use things INSIDE that function to change stuff where you called it, you have only two choices in C.
You can either return the value and assign it to a variable in the calling function like this:
int result = myFunction(someVariable, anotherVariable);
//do stuff with result
Or, use pointers.
I'm not explaining pointers here, that's usually several lectures worth of information, and is one of the more difficult concepts to grasp for introductory programmers. All I can tell you is you need to learn them, but this format is not the right way to go about doing that.

Answer 4

You're passing size to promptData as a copy.
Thus changes to numElem inside promptData will not affect the size variable in your main. Hence size remains uninitialized, i.e. has an undefined value and therefore should not be used as a size for an array.
If you need to initialize an array with a size that's only known at run-time, you need to allocate memory for the array dynamically using malloc, for example:
double* array = malloc(size * sizeof(double));