Copy of address in C?

Question

I know that this code:

void incVar(int i){ i++; }

We know this will create a copy of the integer and then increase that value, but not the actual value.

By nature: Methods in C create copies of parameters in their Stack Frame and not the original variable.

But:

 void incVar(int *i){ (*i)++; }

Is supposed to increase the actual value of the integer by the pointer dereference.

But then, why doesn't C just create a copy of the pointer *i instead? If this is the normal behavior with regular integers, then why doesn't the same thing happen with pointers?

You did not change i so that conclusion just comes out of nowhere. — user555045
– user555045, Commented Jan 17, 2017 at 6:58
It is a common mistake to mix up a pointer declaration int *i with the pointer de-reference operator *i. Stupidly, the C language uses the * symbol for many completely different things. — Lundin
– Lundin, Commented Jan 17, 2017 at 7:46
Possible duplicate of How to understand the pointer star * in C? — Julien Lopez
– Julien Lopez, Commented Jan 17, 2017 at 8:36

StoryTeller - Unslander Monica · Accepted Answer · 2017-01-17 06:56:19Z

It is the same with pointers. All variables in C are passed by value, even pointers.

You copy the address stored in the pointer outside the function, into its parameter.

But you can use that address to reference a variable which can be allocated anywhere. So in the following code:

int j = 0; incVar(&j);

incVar receives by-copy the address of j. But it can use that address to read or modify j (in)directly.

I understand. Thanks! So if I have my first method, the address of the copied x is different from the global x I defined?
@Amad27 - You mean i? Yes, its address is different from the address of the variable that was copied into it. In fact, all valid objects in C have a unique address at any given time.

Michael Karcher · Accepted Answer · 2017-01-17 06:58:37Z

In fact the same thing as with values happens with pointers. Just make sure to understand the syntax correctly. You are not passing the integer *i by copy to incVar, but you are passing the pointer i of type int* by copy. No matter how often you copy the pointer to an address, it always points to the same address. So i in your second incVar example points to the integer the caller took the address of. So by derefencing the copy of the pointer (in (*i)++), you are acessing the integer of the caller.

Lincoln Cheng · Accepted Answer · 2017-01-17 07:52:57Z

Its better to see int *i; as a variable named i of type pointer-to-int, instead of a pointer named *i.

In your example:

 void incVar(int *i) { (*i)++; }

We probably use it as such in the main function:

int a = 5; incVar(&a); //After this line, a is now 6.

What happens in incVar(..) is this:

A variable i is created, of type pointer-to-int.
It holds a copy of the value of &a.
Although i is a copy, its value is the same as that of the value of &a, and it still points to the same integer a.
As a result, by de-referencing i, I refer to the integer a.
Increment the integer a, which obviously is at the address pointed to by i.

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