How to modify an array inside a function in C - arrays

Question

Let's consider the following example :
#include <stdio.h>
void change_byte(int *byte);
int main()
{
int byte = 0;
change_byte(&byte);
printf("Byte : %d \r\n", byte);
}
void change_byte(int *byte)
{
*byte= 5;
}
I am simply changing the value of an integer inside a function by passing the integer as a pointer to the function. It yields :
Byte : 5
Everything's fine.
I want to generalize the function to modify an array of integer instead of an integer. Here is the code :
#include <stdio.h>
#define SIZE_ARRAY 10
void change_array(int *array, int size);
int main()
{
int array[SIZE_ARRAY] = {0};
change_array(array, SIZE_ARRAY);
printf("Array : ");
for(int i = 0 ; i < SIZE_ARRAY ; i++)
{
printf("%d ", array[i]);
}
}
void change_array(int *array, int size)
{
for(int i = 0 ; i < size ; i++)
{
array[i] = 5;
}
}
It yields :
Array : 5 5 5 5 5 5 5 5 5 5
I like it because it does not make use of dynamic allocation, but I have trouble understand how it works. From what I understand, array gets converted into a pointer when entering the function change_array. But when I was changing the value of byte in the previous example, I was doing *byte = 5. Here, I am doing array[i] = 5 and not *array[i] = 5.
Finally, I want to change the previous example to modify array based on a global array :
#include <stdio.h>
#define SIZE_ARRAY 10
int global_array[10] = {5, 5, 5, 5, 5, 5, 5, 5, 5, 5};
void change_array(int *array, int size);
int main()
{
int array[SIZE_ARRAY] = {0};
change_array(array, SIZE_ARRAY);
printf("Array : ");
for(int i = 0 ; i < SIZE_ARRAY ; i++)
{
printf("%d ", array[i]);
}
}
void change_array(int *array, int size)
{
array = global_array;
}
It yields :
Array : 0 0 0 0 0 0 0 0 0 0
Why is it so ? What to I need to change to make it work ?
Thanks.

Answer 1

Consider your first program.
#include <stdio.h>
void change_byte(int *byte);
int main()
{
int byte = 0;
change_byte(&byte);
printf("Byte : %d \r\n", byte);
}
void change_byte(int *byte)
{
*byte= 5;
}
In this program the object byte is passed to the function change_byte by reference through a pointer to it
change_byte(&byte);
In C passing by reference means passing an object indirectly through a pointer tp it.
So dereferencing the pointer byte declared as a function parameter
void change_byte(int *byte);
you get a direct access to the pointed object byte of the type int defined in main.
Now let's consider your second program
#include <stdio.h>
#define SIZE_ARRAY 10
void change_array(int *array, int size);
int main()
{
int array[SIZE_ARRAY] = {0};
change_array(array, SIZE_ARRAY);
printf("Array : ");
for(int i = 0 ; i < SIZE_ARRAY ; i++)
{
printf("%d ", array[i]);
}
}
void change_array(int *array, int size)
{
for(int i = 0 ; i < size ; i++)
{
array[i] = 5;
}
}
In main you declared an integer array
int array[SIZE_ARRAY] = {0};
Array designators used in expressions with rare exceptions are converted to pointers to their first elements.
Thus this call
change_array(array, SIZE_ARRAY);
is equivalent to
change_array( &array[0], SIZE_ARRAY);
So dereferencing the pointer within the function you can change the first element of the array defined in main.
But array elements are stored in a continuous extent of memory. So using the pointer arithmetic and having a pointer to the first element of an array you can access all elements of the array.
In fact all elements of the array array are passed to the function change_array by reference through a pointer to the first element of the array.
For example the for loop within the function you could rewrite like
for(int i = 0 ; i < size ; i++)
{
*( array + i ) = 5;
}
Now let's consider your third program.
#include <stdio.h>
#define SIZE_ARRAY 10
int global_array[10] = {5, 5, 5, 5, 5, 5, 5, 5, 5, 5};
void change_array(int *array, int size);
int main()
{
int array[SIZE_ARRAY] = {0};
change_array(array, SIZE_ARRAY);
printf("Array : ");
for(int i = 0 ; i < SIZE_ARRAY ; i++)
{
printf("%d ", array[i]);
}
}
void change_array(int *array, int size)
{
array = global_array;
}
As it was pointed out already the array array passed to the function change_array is converted to a pointer to its first element.
You may imagine the function call and its definition the following way (I will rename the first function parameter that to avoid name ambiguity).
change_array(array, SIZE_ARRAY);
//...
void change_array( /* int *parm_array, int size */)
{
int * parm_array = array;
int size = SIZE_ARRAY;
parm_array = global_array;
}
That is function parameters are its local variables. The parameter parm_array is alive until the function stops its execution.
Thus this statement
parm_array = global_array;
assign the pointer to the first element of the global array global_array to the local variable parm_array of the function. This assignment does not touch in any way the array array defined in main. It only changes the local variable parm_array declared in the function change_array.
To achieve the expected by you result you could define in main a pointer that is initialized by the array array. And in the function change_array you could reassigned the pointer with the array global_array passing it to the function by reference the same way as you passed the object byte in your first program.
Here is a demonstrative program.
#include <stdio.h>
#define SIZE_ARRAY 10
int global_array[10] = {5, 5, 5, 5, 5, 5, 5, 5, 5, 5};
void change_array( int **array_ptr );
int main()
{
int array[SIZE_ARRAY] = {0};
int *array_ptr = array;
change_array( &array_ptr );
printf("Array : ");
for(int i = 0 ; i < SIZE_ARRAY ; i++)
{
printf("%d ", array_ptr[i]);
}
}
void change_array( int **array_ptr )
{
*array_ptr = global_array;
}
The program output is'
Array : 5 5 5 5 5 5 5 5 5 5

Answer 2

For a better understanding (and exercise), edit your function
void change_array(int *array, int size)
{
array = global_array;
}
to
void change_array(int *array, int size)
{
array = global_array;
for (int i=0; i < size; ++i)
printf("%d ", array[i]);
}
Ask yourself, what the output 'means'.