Pointers: &, *, and pass-by-pointer

A problem from last lesson

• Last lesson, a function got a copy of your number. Changing the copy did not change the original.
• C copies every argument you pass to a function. So a normal function cannot change your variable.
• To change it, the function needs the variable's address — where it lives in memory.

Every variable has an address

• Think of a variable as a box in memory. The value is kept inside the box.
• Every box has an address: a number that says where the box is.
• The & operator gives you that address. &x means "the address of x".

A pointer holds an address

• A pointer is a variable that stores an address, not a normal value.
• You declare one with a *: int *p = &x; reads as "p points to x".
• To reach the value at that address, put * in front: *p. This is called dereference.
• Changing *p changes the original variable. *p = 5; puts 5 into x itself.

#include <stdio.h>

void set_to_five(int *p) {
    *p = 5;            // change the value at the address
}

int main(void) {
    int x = 0;
    set_to_five(&x);   // pass the ADDRESS of x
    printf("%d\n", x); // 5 — the original changed!
    return 0;
}

Out-pointers: giving back more than one answer

• A function can return only one value. But it can change several variables through pointers.
• You pass the address of each answer, like find(&min, &max). The function writes into them with *.
• These extra pointer parameters are called out-pointers. The caller reads the answers after the call.
• A pointer that points to nothing is NULL. Never dereference a NULL pointer.

Now you try

• Each task gives you a function to complete. Use * to read or change the value the pointer points to.
• When a value comes in through a pointer (like int *p), write your answer through it — do not return it.
• Do not write a main — the checker provides one. Press Run, then Check answer.