How to pass an array by reference to a function in C? - c

Question

I am a beginner coder in C.
I have the following code:
int main()
{
struct* Array[malloc(10*sizeOf(struct)];
/* I then fill the struct. (int num,float points)
* I then want to pass this array that I've filled up to a sortList function,
* which then outputs the sorted array back to main().
*
* Then I want to output the array as: "number, points" */
}
struct struct
{
int number;
float points;
}
void sortList()
{
/*sort Array, then return sorted array*/
}
How would I pass the array and then back?
Any links or suggestions are greatly helpful.

Answer 1

How to pass an array by reference ...?
When an array a is passed to a function foo(a), it is the actual argument. In C, when an array is passed to a function, rather than the entire array being given to foo(), the expression a is converted to the address of the first element of the array.
Except when it is the operand of the sizeof operator, the _Alignof operator, or the unary & operator, or is a string literal used to initialize an array, an expression that has type ‘‘array of type’’ is converted to an expression with type ‘‘pointer to type’’ that points to the initial element of the array object and is not an lvalue. ... C11dr §6.3.2.1 3
So foo() is given an address of type int *. Let us assume the value is 1234.
Inside the function void foo(int *fa), the formal parameter fa takes on the value 1234.
From the caller's point-of view, this is pass-by-reference as a is affected by foo() and the function received a "reference". From the function's point-of-view, this is pass-by-value as fa gets a copy of the converted a. In C, it is this second-point-of view that is usually spoken of when folks says that C does not pass anything by reference. In both cases, fa is an int * and not an array.
foo() had the address of main's a in the variable fa. So code fa[0] = 456 sets a value that is still seen after foo() completes.
void foo(int *fa) {
fa[0] = 456;
}
int main(void) {
int a[5];
a[0] = 123;
printf("%d\n", a[0]); // prints 123
foo(a);
printf("%d\n", a[0]); // prints 456
return 0;
}
I used a simply int array to explain things. Yet original code had other problems. Below are ideas on how to allocate memory.
// struct* Array[malloc(10*sizeOf(struct)];
struct ok12type* ok12Array = malloc(sizeof *ok12Array * 10);
....
// Do stuff with ok12Array
...
free(ok12Array);
ok12Array = NULL;
// Do _not_ do stuff with ok12Array

Answer 2

Code seems to be wrong.
In C, arrays are always pass by reference.

Answer 3

First off you should name your struct. I gave it the name typename in my examples below.
To allocate the memory, you want:
struct typename *Array = malloc(10*sizeof(struct typename));
Passing the array to a function is easy, just declare your function as:
void sortList(struct typename *Array)
{
}
You can then work on the array directly in the function, because arrays are always pass by reference, never pass by value.
So your program might look like:
struct typename
{
int number;
float points;
}
int main()
{
struct typename *Array = malloc(10*sizeof(struct typename));
Array[0].number = 5;
Array[0].points = 7;
Array[1].number = 2;
Array[1].points = 11;
/* Fill the rest of the array. */
/* Sort it. */
sortList(Array);
}
void sortList(struct typename *Array)
{
/* Do your sorting here. */
}