pointers on pointers syntax

[HomesickAlien]

Hi all,

I am just reading the C++ tutorial from this site and stumbled upon some understanding problems concerning pointers. On page 71 there is a discussion on "Pointers on Pointers". There is an example with the expression

char ** c;

It is then argued that this can have three meanings (which leads to different values):

1. c has type char **
2. c* has type char *
3. c** has type char

I am not sure if I understand the exact meaning here and therefore cannot understand the example. Could someone please explain exactly how to discern these cases?

[Gaminic]

It's not about "discerning the cases". All three are true, at all times. Read it aloud in words:
**c is a character
*c is a pointer to a character
c is a pointer to a pointer to a character

char** ppc;
char* pc;
char c;

**ppc = *pc; // Assign char to char
**pcc = c; // Idem
*ppc = pc; // Assign pointer to char to pointer to char
*ppc = &c; // Assign pointer to char an address of a char
ppc = &pc; // Assign pointer to pointer to char the address of a pointer to char
ppc = &&c; // This will throw errors because it is meaningless, but "grammatically" it is correct 

[hamsterman]

Is it clear how a simple pointer char* ptr; can have two 'meanings' (uses) :
1. ptr has type char*
2. *ptr has type char
?

This is the same thing, just with one more *. As long as a value is a pointer, you can apply * to it and get what it points to.

[HomesickAlien]

I really don't get it how to arrive at 8092, 7230 and 'z' in the example. Everything before this I could follow easily.

[z05DSL3A]

char a;
char * b;
char ** c;
a = 'z';
b = &a;
c = &b;

+----+    +----+    +----+
| &B |--->| &A |--->| Z  |
+----+    +----+    +----+
  C         B         A

If you look at the value of C it has the address of where B can be found.
If you look at the value of what C is pointing to (ie *C) you go to C find the address of what it is pointing to and get the value of that, in this case it is the address of A.
If you dreference C again (ie **C), you go to c find that address, follow it to B, and follow that address to A and the value of that is the letter Z.

So,
C has the value of &B
*C has the value of &A
**C has the value of Z

[Gaminic]

Found your example. I'm not sure how you can follow the rest if you don't understand this part, but okay.

Think of it this way: a variable is actually just a value saved at a memory address. The internal workings of the system map the variable names to memory addresses (don't ask me how; it doesn't matter anyway).

In this case, there's a char variable ('z') saved at a memory slot ('7230') (the actual number is meaningless; in your debugger, you'll see a hex number that changes at every runtime). Secondly, we have a pointer to a char. It, too, is a value ('7230') saved at a memory slot ('8092'). The only change is that the value is an address rather than a char. Because it is of type *char, we know the value at the address pointed to is to be interpreted as a char. Thirdly, we have a pointer to a pointer to a char. Again, it is a value ('8092') saved at a memory slot ('10502').

Now, if I use variable c, the system knows I'm talking about the value saved at 10502 (internally). Because the compiler knows c is of type char**, it knows that *c (dereferenced once) is a char* and **c (dereferenced twice, equal to *(*c)) is a char.

Basically, pointers are just an easy way to keep track of a variable. If 'a' is ever assigned a new value, any copy we have would be invalid (and we'd have no way of knowing). However, since 'a' will still be at the same memory slot, b and c would still be correct, as they would still be keeping track of 'a', regardless of any changes that happen to it.

[HomesickAlien]

@Grey Wolf and Gaminic
Thanks to you both, I think I get it now. I don't know where the confusion came from. Perhaps one has to do more exercises with this stuff, as usual.

[Gaminic]

Don't worry about pointers. They'll grow on you soon enough!