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From: francus@cheshire.columbia.edu (Yoseff Francus)
Newsgroups: comp.lang.c
Subject: Re: C pointer to a function
Message-ID: <5017@columbia.edu>
Date: Sat, 26-Sep-87 20:33:48 EDT
Article-I.D.: columbia.5017
Posted: Sat Sep 26 20:33:48 1987
Date-Received: Sun, 27-Sep-87 11:38:29 EDT
References: <501@acf3.NYU.EDU>
Sender: nobody@columbia.edu
Reply-To: francus@cheshire.columbia.edu.UUCP (Yoseff Francus)
Distribution: na
Organization: Columbia University CS Department
Lines: 48
Keywords: static pointer function

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




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yf
In Xanadu did Kubla Khan a stately pleasure dome decree
But only if the NFL to a franchise would agree.

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