struct & enum & :: visibility operator
robbio
I have a header file like this:
-------- file.h ------------
....
extern "C"
{
...
typedef _S S;
...
struct _S
{
...
enum
{
a,
b
}myenum;
...
};
}
--------- end of file.h -------------
and a C file like this
----------- file.c ----------------
.....
#include "file.h"
static void myFunc(S* pS)
{
....
if (... == _S::a)
....
....
}
-----------end of file.c --------------
I'm using the Microsoft 2005 environment and so its compiler. The compiler
says me that there is an error in myFunc, the error is that _S is an
undeclared indentifier. I'm not an expert of C but I don't know where the
problem is.
Can anyone help me, please?
thanks a lot
roberto
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Keith Thompson
> Hi at all,
>
> I have a header file like this:
>
> -------- file.h ------------
> ....
> extern "C"
extern "C" is a syntax error in C. Apparently this header is meant to
be C++, which is a different language with its own newsgroups.
It's common to use something like this:
#ifdef __cplusplus
extern "C" {
#endif
/* ... */
#ifdef __cplusplus
}
#endif
if you really feel the need to use the same header for both C and C++.
> {
> ...
> typedef _S S;
This attempts to declare "S" as an alias for the type "_S". You
haven't declared a type called "_S", so of course the compiler doesn't
know what you're talking about. Things don't become visible until
*after* you declare them.
Well, mostly. You *could* have declared
typedef struct _S S;
This introduces "struct _S" as an incomplete type; you can't declare
an object of that type, but you can declare a typedef or a pointer
type that uses it.
> ...
>
> struct _S
And, in C, this doesn't declare a type "_S"; it declares a type
"struct _S". Unlike in C++, you can't refer to "struct _S" as just
"_S".
Also, it's dangerous to use identifiers starting with an underscore;
such identifiers are reserved to the implementation. The rule is
actually a little more complicated, but that's all you need to
remember: don't declare things using identifiers starting with
underscores.
You don't really *need* the typedef at all. You can just declare your
type as:
struct S { /* ... */ };
and refer to it as "struct S". Or, if you prefer to have a one-word
name for the type, you can use the same identifier for the typedef as
for the struct tag:
typedef struct S { /* ... */ } S;
If you need to refer to the type name inside the declaration, things
are a little more complicated; I'll leave the details for later.
--
Keith Thompson (The_Other_Keith) ks...@mib.org <http://www.ghoti.net/~kst>
Nokia
"We must do something. This is something. Therefore, we must do this."
-- Antony Jay and Jonathan Lynn, "Yes Minister"
Barry Schwarz
<robb...@libero.it> wrote:
snip C++ code
You might have better luck asking C++ questions in a C++ discussion
group.
Remove del for email
Kalle Olavi Niemitalo
> extern "C"
> {
C does not support extern "C". Put #ifdef __cplusplus around
that and the braces.
#ifdef __cplusplus
extern "C"
{
#endif
...
#ifdef __cplusplus
}
#endif
> ...
> typedef _S S;
> ...
>
> struct _S
_S is a reserved identifier because it begins with an underscore
and a capital letter. Use some other name, for example S.
Besides, the typedef already must specify that it refers to a
struct type.
typedef struct S S;
...
struct S
{
...
};
That will work in C++ as well.
> static void myFunc(S* pS)
> {
> ....
> if (... == _S::a)
C does not support a :: operator.
Although you defined the enum inside the struct,
C places the constants of the enum in the global scope,
so you can just use a and b here.
if (... == a)
You should therefore choose longer names that are less likely
to conflict.
Hans-Bernhard Bröker
> Hi at all,
>
> I have a header file like this:
>
> -------- file.h ------------
> ....
> extern "C"
> {
As shown, this problem is manifestly off-topic. There's no 'extern "C"'
in the C programming language.
Your problem is with C++, not C. Yes, that's a different language.