Function arguments by reference

#Tutorial12#


Function arguments are passed by value, which means they are copied in and out of functions. But what if we copied pointers to values instead of the values themselves? This will enable us to give functions control over variables and structures of the parent functions, and not just a copy of them.
Let's say we want to write a function which increments a number by one, called addone. This will not work:

void addone(int n) {
    n++;
}

int n;
printf("Before: %d\n", n);
addone(n);
printf("After: %d\n", n);

Execute Code However, this will work:

void addone(int * n) {
    (*n)++;
}

int n;
printf("Before: %d\n", n);
addone(&n);
printf("After: %d\n", n);

The difference is that the second version of addone receives a pointer to the variable n as an argument, and then it can manipulate it, because it knows where it is in the memory.
Notice that when calling the addone function, we must pass a reference to the variable n, and not the variable itself - this is done so that the function knows the address of the variable, and won't just receive a copy of the variable itself.

Pointers to structures

Let's say we want to create a function which moves a point forward in both x and y directions, called move. Instead of sending two pointers, we can now send only one pointer to the function of the point structure:

void move(point * p) {
    (*p).x++;
    (*p).y++;
}

Execute Code However, if we wish to dereference a structure and access one of it's internal members, we have a shorthand syntax for that, because this operation is widely used in data structures. We can rewrite this function using the following syntax:

void move(point * p) {
    p->x++;
    p->y++;
}