address of pointer from a pointer

Hi All,

I can't seem to wrap my head around this one.

I have a pointer,

int *x;

which I can assign:

x = &y;

Now, I can send this variable by reference like so:

some_function( &x );

But, what do I do if x is contained within a pointer to a struct (or class
for that matter)?

struct c {
int *x;
}

such that a pointer to c (c *cptr) would access it like this:

cptr->x = &y;

but now, if I try to send this by reference:

some_function( cptr->&x )

I get an error: expected unqualified-id before '&' token. How do I
properly send this variable by reference?

Thanks for any help!
Scott

re: address of pointer from a pointer

Scott wrote:[color=blue]
> I can't seem to wrap my head around this one.
>
> I have a pointer,
>
> int *x;
>
> which I can assign:
>
> x = &y;
>
> Now, I can send this variable by reference like so:
>
> some_function( &x );
>
> But, what do I do if x is contained within a pointer to a struct (or
> class for that matter)?
>
> struct c {
> int *x;
> }
>
> such that a pointer to c (c *cptr) would access it like this:
>
> cptr->x = &y;
>
> but now, if I try to send this by reference:
>
> some_function( cptr->&x )
>
> I get an error: expected unqualified-id before '&' token. How do I
> properly send this variable by reference?[/color]

The variable is (cptr->x). The address of it is...? (&(cptr->x)).
Use parentheses to fully contain (limit, denote, delineate) your
expression. Then add to your expression. Only remove parentheses
when you're ready and know the rules of precedence.

V
--
Please remove capital As from my address when replying by mail

re: address of pointer from a pointer

Scott wrote:[color=blue]
> Hi All,[/color]

Howdy-doo.
[color=blue]
> I have a pointer,
>
> int *x;[/color]

Okay.
[color=blue]
> which I can assign:
>
> x = &y;[/color]

Yes, assuming y is of type int.
[color=blue]
> Now, I can send this variable by reference like so:
>
> some_function( &x );[/color]

Yes, assuming that some_function takes one argument of type int*. I
assume when you refer to passing "by reference," you mean "by pointer."
Remember, C++ has a different construct which is actually *named*
"reference," and that's a common way to pass parameters as well --
considering that, it's maybe better not to say "pass by reference" when
you mean you're using a pointer, if there's any chance of confusion.
And since the address-of operator (&) uses the same symbol as that used
to declare a reference type, I'd say confusion can crop up pretty
easily here.
[color=blue]
> But, what do I do if x is contained within a pointer to a struct (or class
> for that matter)?
>
> struct c {
> int *x;
> }
>
> such that a pointer to c (c *cptr) would access it like this:
>
> cptr->x = &y;[/color]

No problem. It's a matter of scoping. In C++, as in many other
languages, we have to deal with scoping a lot. When referring to
something that doesn't reside in your local scope, you have to qualify
it. So, you can't just say "x" (as you know); you have to say cptr->x
(or (*cptr).x, which is the same thing). This is true whenever you
refer to x.
[color=blue]
> but now, if I try to send this by reference:
>
> some_function( cptr->&x )
>
> I get an error: expected unqualified-id before '&' token.[/color]

Right. The address-of operator is looking to the right of itself for a
symbol which names a variable to take the address of. The variable
name has to be qualified because it's not local, so you want:
some_function(&(cptr->x))

The compiler is actually balking slightly before it gets a chance to
consider the address-of operator, though, and that's what the error
message is about. The message is saying that the arrow operator has to
point to an unqualified-id -- that is, a variable name which is not
missing any necessary qualifiers (namespaces, enclosing classes, etc.).
Instead, it sees the symbol '&', which is not valid.
[color=blue]
> How do I
> properly send this variable by reference?[/color]

See above.
[color=blue]
> Thanks for any help!
> Scott[/color]

HTH,
Luke

re: address of pointer from a pointer

>> I have a pointer,[color=blue][color=green]
>>
>> int *x;
>> which I can assign:
>>
>> x = &y;
>> Now, I can send this variable by reference like so:
>>
>> some_function( &x );[/color]
>
> Yes, assuming that some_function takes one argument of type int*.[/color]

Warning W8069 solution.c 7: Nonportable pointer conversion in function f
Warning W8075 solution.c 14: Suspicious pointer conversion in function main

The function's signature should be: some_function( int ** ). However,
some_function( int *) seems to produce the same results, even though you are
warned by the compiler. Is there anything that can go wrong when you pass
the address of a pointer to a function that does not recieve a pointer to a
pointer?

Thx

re: address of pointer from a pointer

"jimjim" <netuser@blueyonder.co.uk> wrote in message
news:Xi9Vf.41750$wl.21461@text.news.blueyonder.co. uk...[color=blue][color=green][color=darkred]
>>> I have a pointer,
>>>
>>> int *x;
>>> which I can assign:
>>>
>>> x = &y;
>>> Now, I can send this variable by reference like so:
>>>
>>> some_function( &x );[/color]
>>
>> Yes, assuming that some_function takes one argument of type int*.[/color]
>
> Warning W8069 solution.c 7: Nonportable pointer conversion in function f
> Warning W8075 solution.c 14: Suspicious pointer conversion in function
> main
>
> The function's signature should be: some_function( int ** ). However,
> some_function( int *) seems to produce the same results, even though you
> are warned by the compiler. Is there anything that can go wrong when you
> pass the address of a pointer to a function that does not recieve a
> pointer to a pointer?
>
> Thx[/color]

Oups, my mistake! The C++ compiler outputs:

Error E2034 solution.cpp 14: Cannot convert 'int * *' to 'int *' in function
main()
Error E2342 solution.cpp 14: Type mismatch in parameter 'ip' (wanted 'int
*', got 'int * *') in function main()

I made the mistake and gave a '.c' extension to my file, and therefore it
was compiled with the C compiler. Apparently, C is more tolerant to such
syntax.

However, could anyone attempt to answer my question, which of course is in
the context of the C lang?

Thx in advance.

jimjim

re: address of pointer from a pointer

Hi Luke,

On Fri, 24 Mar 2006 20:54:18 -0800, Luke Meyers wrote:[color=blue]
> Right. The address-of operator is looking to the right of itself for a
> symbol which names a variable to take the address of. The variable
> name has to be qualified because it's not local, so you want:
> some_function(&(cptr->x))
>
> The compiler is actually balking slightly before it gets a chance to
> consider the address-of operator, though, and that's what the error
> message is about. The message is saying that the arrow operator has to
> point to an unqualified-id -- that is, a variable name which is not
> missing any necessary qualifiers (namespaces, enclosing classes, etc.).
> Instead, it sees the symbol '&', which is not valid.[/color]

Excellent... thanks for the help!

Scott

re: address of pointer from a pointer

jimjim wrote:[color=blue][color=green][color=darkred]
> >> I have a pointer,
> >>
> >> int *x;
> >> which I can assign:
> >>
> >> x = &y;
> >> Now, I can send this variable by reference like so:
> >>
> >> some_function( &x );[/color]
> >
> > Yes, assuming that some_function takes one argument of type int*.[/color]
>
> Warning W8069 solution.c 7: Nonportable pointer conversion in function f
> Warning W8075 solution.c 14: Suspicious pointer conversion in function main[/color]

You have to post the code (all of it); nobody has the time to try and
guess.
[color=blue]
> The function's signature should be: some_function( int ** ).[/color]

Right, yes -- I think I misread it the first time. x is of type int*,
so &x is of type int**.
[color=blue]
> However,
> some_function( int *) seems to produce the same results, even though you are
> warned by the compiler.[/color]

I'd treat that warning as an error. I'm surprised it isn't one.
[color=blue]
> Is there anything that can go wrong when you pass
> the address of a pointer to a function that does not recieve a pointer to a
> pointer?[/color]

You should get an error, or a warning that should be taken as an error.
That's what the type system is for -- to make sure you don't (among
other things) pass parameters of the wrong type.

Anyway -- if you still have questions, post the code, THEN ask.

Luke