From: mrscreen AT hotmail DOT com (Heliophobe)
Newsgroups: comp.os.msdos.djgpp
Subject: Re: Passing/updating pointers...
Date: Sat, 13 Mar 1999 12:15:41 -0800
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References: <3 DOT 0 DOT 6 DOT 16 DOT 19990313015041 DOT 1dbffe6a AT pop DOT detroit DOT crosswinds DOT net>
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To: djgpp AT delorie DOT com
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Reply-To: djgpp AT delorie DOT com

In article <3 DOT 0 DOT 6 DOT 16 DOT 19990313015041 DOT 1dbffe6a AT pop DOT detroit DOT crosswinds DOT net>,
djgpp AT delorie DOT com wrote:

>I have never quite figured out how to pass a pointer to a function and have
>the function update the variable sent to it.  So, for many years I have
>just found ways to work around this little glitch in my memory.  Now, I
>have come to a point in a program that I'm writing where I can't work
>around the problem anymore.

[snip]

>So, how would I go about doing this?  That is, how would I call the
>function and how would I perform the update?

Just passing a pointer to a function [void function(char *input)] gives
the address the pointer points to, it doesn't tell the function where the
pointer itself is, so it doesn't know how to change it.


If you want to have a function update not just the data a pointer points
to, but the address it points to, you need to pass the address of the
pointer. That address points to the pointer. Here's a short program
demonstrating this.

#include <stdio.h>

void change_pointer(char **in) /*This means it expects a pointer to a char
pointer.*/
{
 *in+=6;  /*update the pointer that in points to.*/
 return;
}

main()
{
 char text[]="Don't Edit Me!\n";
 char *text_pointer;
 text_pointer=text;   
 printf("%s",text_pointer);  /*before....*/
 change_pointer(&text_pointer);   /*That means, pass the address of the
pointer, NOT the address of the text*/
 printf("%s",text_pointer);  /*After*/
 return(0);
}

Output;

Don't Edit Me!
Edit Me!


The actual text isn't changed, text_pointer just points to a different
part of the text.


Does this answer your question?