A #define that I´m not used to

[Morris Dovey]

Hello all,

Bom dia!

Could someone have the generosity of explaining me what this #defines
define?

#define YPAD_PROTOTYPE(PAD) extern void PAD ();

Expands to produce a prototype statement. For example:

YPAD_PROTOTYPE(my_function)

produces

extern void my_function ();

#define YPAD_FUNCTION(PAD) extern void PAD (VarP) xPrsNode VarP;

Here

YPAD_FUNCTION(my_function)

Expands to produce

extern void my_function (VarP) xPrsNode VarP

which is almost certainly expanded further by other macro
definitions.

 

[alsmeirelles]

Hello all,

Could someone have the generosity of explaining me what this #defines
define?

#define YPAD_PROTOTYPE(PAD) extern void PAD ();
#define YPAD_FUNCTION(PAD) extern void PAD (VarP) xPrsNode VarP;
#define YPRD_PROTOTYPE(PAD) extern int PAD ();
#define YPRD_FUNCTION(PAD) extern int PAD (VarP) xPrsNode VarP;


Thanks a lot,

André LS Meirelles

 

[alsmeirelles]

Bom dia!

You got it right and clear portuguese!

Expands to produce a prototype statement. For example:

YPAD_PROTOTYPE(my_function)

produces

extern void my_function ();


Here

YPAD_FUNCTION(my_function)

Expands to produce

extern void my_function (VarP) xPrsNode VarP

which is almost certainly expanded further by other macro
definitions.

Yeah, that makes a lot of sense, thanks!!

 

[Eric Sosman]

Bom dia!


Expands to produce a prototype statement. For example:

YPAD_PROTOTYPE(my_function)

produces

extern void my_function ();

Yes, but this is not "a prototype statement." It is a
function declaration with no prototype.

Here

YPAD_FUNCTION(my_function)

Expands to produce

extern void my_function (VarP) xPrsNode VarP

No, there is also a semicolon `;' in the macro expansion.

which is almost certainly expanded further by other macro
definitions.

It may be further expanded, but I wouldn't say "almost
certainly." It looks very much like the start of an old-
style (pre-prototype) function definition, and is probably
followed by a `{}'-enclosed block that is the function body.

The `extern' is unusual, but harmless.

 

[alsmeirelles]

Yes, but this is not "a prototype statement." It is a
function declaration with no prototype.



No, there is also a semicolon `;' in the macro expansion.


It may be further expanded, but I wouldn't say "almost
certainly." It looks very much like the start of an old-
style (pre-prototype) function definition, and is probably
followed by a `{}'-enclosed block that is the function body.

That´s the problem, I don´t know wich #define is beeing expanded. The
function has a body (a big one in fact). The definition starts like
this:

YPAD_FUNCTION(yPAD_mmc)

 

[Eric Sosman]

[...]
That´s the problem, I don´t know wich #define is beeing expanded. The
function has a body (a big one in fact). The definition starts like
this:

YPAD_FUNCTION(yPAD_mmc)

The expansion is

extern void yPAD_mmc (VarP) xPrsNode VarP;

Have you, perhaps, been confused by the similarity of
the macro names? They differ by just one character, buried
in the middle and topologically similar:

#define YPAD_PROTOTYPE ...
#define YPRD_PROTOTYPE ...
^

#define YPAD_FUNCTION ...
#define YPRD_FUNCTION ...
^

 

[Kaz Kylheku]

Here

   YPAD_FUNCTION(my_function)

Expands to produce

   extern void my_function (VarP) xPrsNode VarP

Also, there is a semicolon.

which is almost certainly expanded further by other macro
definitions.

I would guess that likely works like this:

YPAD_FUNCTION(my_function)
{
/* body that references VarP of type xPrsNode
}

My guess is that the YPAD macros are conditionally defined to switch
between ANSI declarations and old-style.