References and Dereferences

Pointers are built around two operators:

The address-of operator &.
The dereference operator *.

Address-of Operator

The address-of operator, represented by &, allows us to directly interact with our computer's RAM. Using & lets you obtain the actual memory address of an object.

Main

#include <stdio.h>

int main()

{

int x = 100;

printf("Value of variable: %d\n", x);

printf("Address of variable into RAM: %p\n", &x); // using reference operator `&` for getting address

return 0;

}


              123456789101112
            
#include <stdio.h>

int main()
{
    int x = 100;

    printf("Value of variable: %d\n", x);

    printf("Address of variable into RAM: %p\n", &x); // using reference operator `&` for getting address

    return 0;
}

Note

%p is the format specifier used for addresses (pointer).

Addresses are typically expressed in hexadecimal notation.

Think of the & operator as identifying your home's address using your name.

Dereference Operator

Conversely, the * operator gives you the resident's name when given their address. So, how can we employ this operator if we're not directly dealing with addresses? If you have an expression like &x, which returns the address of the x variable, applying the * operator to it (*&x) gives you the value of the variable stored at that address.

Note

Essentially, *&x is the same as x.

Main

#include <stdio.h>

int main()

{

int x = 100;

printf("Value of variable: %d\n", x);

printf("Address of variable in RAM: %p\n", &x); // using reference operator `&`

printf("Dereferencing address of variable: %d", *(&x)); // using dereference operator `*`

return 0;

}


              1234567891011121314
            
#include <stdio.h>

int main()
{
    int x = 100;

    printf("Value of variable: %d\n", x);

    printf("Address of variable in RAM: %p\n", &x); // using reference operator `&`

    printf("Dereferencing address of variable: %d", *(&x)); // using dereference operator `*`

    return 0;
}

Note

Don't mix up the dereference operator (*x) with the multiplication operator (x*y).

Opgave

Swipe to start coding

Create an integer array of 5 elements and populate it.
Retrieve the address of the third element.
Increment the address of the third element (i.e., address + 1).
Attempt to dereference the address obtained in the previous step.

Løsning

#include <stdio.h>

int main()

{

int array[] = { 1, 2,3, 4, 5 };

printf("Value of third element %d\n", array[2]);

printf("Address of third element %p\n", &(array[2]));

printf("(address-of-third-element) + 1: %d\n", *((&array[2]) + 1));

return 0;

}

Var alt klart?

Tak for dine kommentarer!

Sektion 6. Kapitel 2

#include <stdio.h>

int main()

{

int array[] = { ___, ___, ___, ___, ___ };

printf("Value of third element %d\n", array[2]);

printf("Address of third element %p\n", ___(array[2]));

printf("(address-of-third-element) + 1: %d\n", ___((___array[2]) + 1));

return 0;

}