How to understand function prototype signal()

[None]

Hello,

1.
The prototype of function signal in signal.h:

void (*signal(int sig, void (*func)(int)))( int);


is some complex to me. Would you please explain it to me in detail with
the C language syntax itself. Thank you!



2.
And I saw another function prototype:

void interrupt ( *oldhandler)();


-Is it legal?

-If it's a right prototype, what does identifier "interrupt" mean? Will
it be a type qualifier.



---
Thanks
lovecreatesbeau ty

[S.Tobias]

Re: How to understand function prototype signal()

None <lovecreatesbea uty@gmail.com> wrote:

[color=blue]
> void (*signal(int sig, void (*func)(int)))( int);[/color]

void (*)(int)

type pointer to (function with one int parameter returning void)

*signal()

signal is a function returning pointer
(Note:
(*signal)()
would mean: signal is a pointer to function)

void (*signal())(int );

signal is a function returning pointer to (function with one int
param returning void)

void (*signal(int sig, void (*func)(int)))( int);

signal is a function with two parameters, returning pointer etc...
The first parameter is type int, the second parameter is type:

void (*)(int)

ie. pointer to function with one int parameter returning void.


The original declaration is equivalent to:
void (*signal(int, void (*)(int)))(int) ;
ie. the parameters of `signal' don't have to be named, similarly
parameter to `func' was not named.

Note that the type of the second parameter of `signal' is the same
as the type that `signal' returns, ie. "pointer to (fn with one int
param, returning void)".

It's easier to understand if you declare a type:

typedef void (*ptr_void_fn_i nt)(int);
ptr_void_fn_int signal(int, ptr_void_fn_int );

(ptr_void_fn_in t is type "pointer to (fn with one etc...)").

`signal' function for a given signal number sets the new signal
handler you supply (which is type ptr_void_fn_int ) and returns
you the previous handler (which naturally has to be the same type).
(For exact `signal' semantics see N869 7.14.1.1.)

[color=blue]
> void interrupt ( *oldhandler)();[/color]
[color=blue]
> -Is it legal?[/color]
[color=blue]
> -If it's a right prototype, what does identifier "interrupt" mean? Will
> it be a type qualifier.[/color]

`interrupt' is not an ISO C keyword. It probably denotes function
call convention, and means that the function `oldhandler' points to
will be called from outside as an interrupt handler (but I might be
very wrong here). You have to consult your compiler manual.

--
Stan Tobias
mailx `echo siXtY@FamOuS.Be dBuG.pAlS.INVALID | sed s/[[:upper:]]//g`

[lovecreatesbeauty]

Re: How to understand function prototype signal()

Stan Tobias,

Thank you

S.Tobias wrote:[color=blue]
> None <lovecreatesbea uty@gmail.com> wrote:
>
>[color=green]
> > void (*signal(int sig, void (*func)(int)))( int);[/color]
>
> void (*)(int)
>
> type pointer to (function with one int parameter returning void)
>
> *signal()
>
> signal is a function returning pointer
> (Note:
> (*signal)()
> would mean: signal is a pointer to function)
>
> void (*signal())(int );
>
> signal is a function returning pointer to (function with one int
> param returning void)
>
> void (*signal(int sig, void (*func)(int)))( int);
>
> signal is a function with two parameters, returning pointer etc...
> The first parameter is type int, the second parameter is type:
>
> void (*)(int)
>
> ie. pointer to function with one int parameter returning void.
>
>
> The original declaration is equivalent to:
> void (*signal(int, void (*)(int)))(int) ;
> ie. the parameters of `signal' don't have to be named, similarly
> parameter to `func' was not named.
>
> Note that the type of the second parameter of `signal' is the same
> as the type that `signal' returns, ie. "pointer to (fn with one int
> param, returning void)".
>
> It's easier to understand if you declare a type:
>
> typedef void (*ptr_void_fn_i nt)(int);
> ptr_void_fn_int signal(int, ptr_void_fn_int );
>
> (ptr_void_fn_in t is type "pointer to (fn with one etc...)").
>
> `signal' function for a given signal number sets the new signal
> handler you supply (which is type ptr_void_fn_int ) and returns
> you the previous handler (which naturally has to be the same type).
> (For exact `signal' semantics see N869 7.14.1.1.)
>
>[color=green]
> > void interrupt ( *oldhandler)();[/color]
>[color=green]
> > -Is it legal?[/color]
>[color=green]
> > -If it's a right prototype, what does identifier "interrupt" mean?[/color][/color]
Will[color=blue][color=green]
> > it be a type qualifier.[/color]
>
> `interrupt' is not an ISO C keyword. It probably denotes function
> call convention, and means that the function `oldhandler' points to
> will be called from outside as an interrupt handler (but I might be
> very wrong here). You have to consult your compiler manual.
>
> --
> Stan Tobias
> mailx `echo siXtY@FamOuS.Be dBuG.pAlS.INVALID | sed s/[[:upper:]]//g`[/color]