Can anybody resolve my mistake - c

Question

I cant seem to find out where i'm going wrong in C. its on line 37 it says assignment to expression with array type any help or advice would be great thanks.
I was wondering also is it something to do with not adding in the brackets to show that they're arrays on line 37 but when i put them in it displays more errors
/*
This program uses pass by reference to calculate the values after two arrays are multiplied by each other
16/02/2015
Jake Young
*/
#include <stdio.h>
#define size 5
//Prototype
int multiply_function(int *[], int *[]);
main()
{
int array1[size];
int array2[size];
int i;
int answer[size];
//get users input for array1
printf("Please enter %d values into array1:\n", size);
for(i=0; i<size; i++)
{
scanf("%d", &array1[i]);
}//end for loop
//get users input for array2
printf("Please enter %d values into array2:\n", size);
for(i=0; i<size; i++)
{
scanf("%d", &array2[i]);
}//end for loop
//call function()
answer=multiply_function(&array1, &array2); // line 37
//Print out the results from array1 multiplied by array2
printf("Array1 multiplied by Array2 is the following:\n");
for(i=0; i<size; i++)
{
printf("%d multiplied by %d is %d\n", array1[i], array2[i], answer[i]);
}//end for loop
}//end main()
multiply_function(int *array1[], int *array2[])
{
int *answer[size];
int i;
for(i=0; i<size; i++)
{
//calculate multiplication
*answer[i]= *array1[i]* *array2[i];
}//end for loop
return(*answer);
}//end function()

Answer 1

int multiply_function(int *[], int *[]);
This doesn't make any sense. You intend to pass arrays of integers to the function, not arrays of pointers. You'll have to study how arrays should be passed to functions.
main()
This form is not standard. Unless you are programming a "bare metal" embedded system, you should use int main (void).
answer=multiply_function(&array1, &array2);
This doesn't make any sense. You declared the function to return an int. Again, study how arrays are passed to and from a function. Furthermore, you can't copy arrays with the assignment operator: you have to use memcpy() or similar functions.
multiply_function(int *array1[], int *array2[])
The function definition is different than the prototype: that is always bad practice. Apart from that, the function doesn't make any sense, as already mentioned.
int *answer[size];
This doesn't make any sense, you are declaring an array of pointers where you want an array of integers.
return(*answer);
Returning a pointer to a local variable in C is always a bug. And you can't return arrays like this. And there is no need for the parenthesis.

Answer 2

OK, you should really invest some more time to study arrays, pointers and fundamentals of functions in C.
Apart from grammatical problems in the code, the fundamental problem in this code is the answer[] array. it is defined both in main() and the multiply_function(). What you must do is to pass this array to the multiply_function() and have the function fill in the array.
I'm giving the solution below, with the hope that you'll compare it to your version and study the differences and continue to learn the basics of C:
#include <stdio.h>
#define size 5
//Prototype
int multiply_function(int *, int *, int *);
main()
{
int array1[size];
int array2[size];
int i;
int answer[size];
//get users input for array1
printf("Please enter %d values into array1:\n", size);
for(i=0; i<size; i++)
{
scanf("%d", &array1[i]);
}//end for loop
//get users input for array2
printf("Please enter %d values into array2:\n", size);
for(i=0; i<size; i++)
{
scanf("%d", &array2[i]);
}//end for loop
//call function()
multiply_function(array1, array2, answer);
//Print out the results from array1 multiplied by array2
printf("Array1 multiplied by Array2 is the following:\n");
for(i=0; i<size; i++)
{
printf("%d multiplied by %d is %d\n", array1[i], array2[i], answer[i]);
}//end for loop
}//end main()
multiply_function(int *array1, int *array2, int *answer)
{
int i;
for(i=0; i<size; i++)
{
//calculate multiplication
answer[i]= array1[i] * array2[i];
}//end for loop
return(*answer);
}//end function()