call by pointers and call by value C - c

Question

How to write this C program pass by pointers and pass by value? I created this program that generates a 10x10 array from -100 to 100 and the main diagonal is read from bottom to top, but I do not know how to create two other programs that are pass by pointers and value
I have to write two more programs with the same input and output only through these two methods. I am a beginner in C and I do not know how to make this.
#include <stdio.h>
#include <stdlib.h>
#define N 10
#define M 10
int my_rand(int max, int min)
{
return (min + rand() / (RAND_MAX / (max - min + 1) + 1));
}
void generate_array(int(*array)[M])
{
printf("Table:");
for (int i = 0; i < N; i++)
{
printf("\n");
for (int j = 0; j < M; j++)
{
array[i][j] = my_rand(-100, 100);
printf("%4d ", array[i][j]);
}
}
}
void print_diagonal(int(*array)[M])
{
printf("\n\nThe main diagonal read from bottom to top:\n");
for (int i = N - 1; i >= 0; i--)
{
printf("%4d ", array[i][i]);
}
}
int *create_array(int(*array)[M])
{
static int new_array[M] = { 0 };
for (int i = 0; i < N; i++)
new_array[i] = array[i][i];
return new_array;
}
void reverseArray(int arr[], int start, int end)
{
int temp;
while (start < end)
{
temp = arr[start];
arr[start] = arr[end];
arr[end] = temp;
start++;
end--;
}
}
void printArray(int arr[])
{
int i;
for (i = N; i < N; i++)
printf("%4d ", arr[i]);
printf("\n");
}
int main(void)
{
int array1[N][M] = { 0 }, *aux_array;
generate_array(array1);
print_diagonal(array1);
aux_array = create_array(array1);
reverseArray(aux_array, 0, N - 1);
printArray(aux_array);
}
Any help would be much appreciated.

Answer 1

C only supports pass by value. The formal argument in the function definition is always a different object in memory from the actual argument in the function call (assuming the actual argument is itself an object, which it doesn't have to be).
We can fake pass-by-reference semantics by passing a pointer (which is passed by value). Here's a typical example:
void swap( int *a, int *b )
{
int tmp = *a;
*a = *b;
*b = tmp;
}
Instead of receiving to integer objects, we receive pointers to two integer objects. We use the unary * operator to dereference the pointers and access the pointed-to objects. If we call that function like so:
int main( void )
{
int x = 1;
int y = 2;
...
swap( &x, &y );
...
return 0;
}
Given the above code, the following are true when we enter the swap function:
a == &x // int *
*a == x == 1 // int
b == &y // int *
*b == y == 2 // int
After we execute the body of the swap function, the following are true:
a == &x // no change
*a == x == 2
b == &y // no change
*b == y == 1
Where things get weird is when you pass array expressions as function arguments. C has a rule that when an array expression is not the operand of the sizeof or unary & operators, or is not a string literal used to initialize a character array in a declaration, the expression is converted ("decays") from an array type to a pointer type, and the value of the expression is the address of the first element.
IOW, given the code
int arr[10];
foo( arr );
Since it's not the operand of the sizeof or unary & operators, the expression arr in the call to foo is automatically converted from type "10-element array of int" to "pointer to int", and the value that gets passed to foo is the address of the first element; IOW, that call is identical to writing
foo( &arr[0] );
and the function definition would be written as
void foo( int *a ) // or int a[], which means the same thing
{
// do something with a[i]
}
IOW, you cannot pass arrays by value in C.
So, looking at your code:
int my_rand(int max, int min)
Both max and min are being passed by value
void generate_array(int(*array)[M])
void print_diagonal(int(*array)[M])
These are receiving a pointer to the first element of a 2D array, so we're faking pass-by-reference.
void reverseArray(int arr[], int start, int end)
We're receiving a pointer to first element of the array (T a[] and T a[N] are both interpreted as T *a in a function parameter declaration), start and end are being passed by value.
void printArray(int arr[])
Again, we're receiving a pointer to the first element of the array.