Question

This is basically just to help me understand pointers better, so if you guys can confirm/deny/explain anything it looks like I don’t understand properly I would be most appreciative. The examples of using mailboxes, and aunts, and streets, and all that crap is just confusing.

int a = 5;

int b = &a; // b will be the memory address of 'a'

int *c = a; // c will be the value of 'a' which is 5

int *d = &a; // d will be a pointer to the memory address of 'a'

int &e = a; // what would this be?


void FunctionA()
{
     int a = 20;
     FunctionB(&a);
     // a is now 15?
}

void FunctionB(int *a)
{
     a = 15;
}

Thank you guys for any help, I am just trying to improve my understanding beyond all of the crappy metaphor explanations im reading.

Answer 1

I’ll take things one by one:

int b = &a; // b will be the memory address of 'a'

No. The compiler (probably) won’t allow this. You’ve defined b to be an int, but &a is the address of an int, so the initialization won’t work.

int *c = a;

No — same problem, but in reverse. You’ve defined c to be a pointer to an int, but you’re trying to initialize it with the value of an int.

int *d = &a;

Yes — you’ve defined d to be a pointer to an int, and you’re assigning the address of an int to it — that’s fine. The address of an int (or an array of ints) is what a pointer to int holds.

int &e = a;

This defines e to be a reference to an int and initializes it as a reference to a. It’s perfectly legitimate, but probably not very useful. Just for Reference, the most common use of a reference is as a function parameter (though there are other purposes, of course).

void FunctionA() { int a = 20; FunctionB(&a); }
void FunctionB(int *a) { a = 15; }

To make this work, you need to change the assignment in FunctionB:

void FunctionB(int *a) { *a = 15; }

As it was, you were trying to assign an int to a pointer, which won’t work. You need to assign the int to the int that the pointer points at to change the value in the calling function.