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From: ftw@datacube.UUCP
Newsgroups: comp.lang.c
Subject: Re: C pointer to a function
Message-ID: <102600017@datacube>
Date: Mon, 28-Sep-87 09:59:00 EDT
Article-I.D.: datacube.102600017
Posted: Mon Sep 28 09:59:00 1987
Date-Received: Wed, 30-Sep-87 05:49:18 EDT
References: <501@acf3.NYU.EDU>
Lines: 65
Nf-ID: #R:acf3.NYU.EDU:-50100:datacube:102600017:000:2371
Nf-From: datacube.UUCP!ftw    Sep 28 09:59:00 1987


> francus@cheshire.columbia.edu.UUCP writes:
> In article <501@acf3.NYU.EDU> hazzah@acf3.NYU.EDU (Ali Hazzah) writes:
> >	Consider the function foo, defined as:
> >
> >	static char *foo(name)
> >	char *name;
> >	{
> >		static char path[20];
> >		char *strcat();
> >		...
> >		return(strcat(path, name);
> >	}
> >
> >	and invoked in the following manner:
> >
> >	callfoo()
> >	char *name;
> >	{
> >	char *path, *foo();
> >	path = foo(name);
> >	}
> >	The question here concerns the scope of the pointer to
> >	the function foo.

"foo" is not a pointer to a function, it is a function returning a pointer.


> >	Specifically, why might the pointer be defined as static?

This is a fine point:  What foo is is a static function that returns a pointer
to a char, not a "function returning a static pointer".  What "static" means
when applied to a function definition like that is that the scope of the
function is only within that source code file.  If I have "foo.c" and "bar.c",
and foo.c has a "static foo()" in it, I cannot call the function foo() from
bar.c

> >	(An example of this can be found on p.23 of Marc Rochkind's
> >	book, "Advanced Unix Programming", 4th printing.)
> 
> The reason its declared static is this: What you are passing back from
> the function is a memory location. The variable that you hope this
> location is a pointer to is the variable path in the function. This
> variable is declared locally in the function. Therefore, when you return
> to callfoo(), there is no guarantee that there will be any reasonable 
> data in the memory location that has been returned to you.  However, since
> its been declared as static the data has to remain at that memory location
> since static guarantees that if you go back into foo(), the data pointed to
> by path will still be there.  Since you are accessing the memory location
> of path (from foo()), and you have declared it static, callfoo() will be
> able to get at the value properly.
> 

The above explains why the array "path" in the function foo is declared as
static.

> 
> ARPA: francus@cs.columbia.edu
> UUCP: seismo!columbia!francus
				Farrell T. Woods 

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