regarding pointers to characters

[sam_cit@yahoo.co.in]

Hi Everyone,

I have the following two cases,

typedef char* cptr;
const cptr ch;
//here ch is a pointer to a character and the pointer is a constant
here...

where as

#define CPTR char*
const CPTR ch;
//here ch is a pointer to character and the character is a constant
here...

What is the reason behind this???

[Keith Thompson]

Re: regarding pointers to characters

sam_cit@yahoo.c o.in writes:

Hi Everyone,
>
I have the following two cases,
>
typedef char* cptr;
const cptr ch;
//here ch is a pointer to a character and the pointer is a constant
here...
>
where as
>
#define CPTR char*
const CPTR ch;
//here ch is a pointer to character and the character is a constant
here...
>
What is the reason behind this???

Is this a homework problem?

--
Keith Thompson (The_Other_Keit h) kst-u@mib.org <http://www.ghoti.net/~kst>
San Diego Supercomputer Center <* <http://users.sdsc.edu/~kst>
We must do something. This is something. Therefore, we must do this.

[sam_cit@yahoo.co.in]

Re: regarding pointers to characters

>

Is this a homework problem?
>

No, not at all, homeworks were a long time ago, i know that macro
(string replacement) is before compilation, but i'm not very clear with
typedefs, i thought answer to this might help me understand it better...

[Keith Thompson]

Re: regarding pointers to characters

sam_cit@yahoo.c o.in writes:

I have the following two cases,
>
typedef char* cptr;
const cptr ch;
//here ch is a pointer to a character and the pointer is a constant
here...
>
where as
>
#define CPTR char*
const CPTR ch;
//here ch is a pointer to character and the character is a constant
here...
>
What is the reason behind this???

A typedef creates an alias for a type. Macro expansion works on text
(tokens, actually) with no regard for what those tokens might mean in
context.

In your second example, "CPTR" expands to "char*", so the declaration
is:

const char* ch;

Consider also the following example:

#include <stdio.h>

#define SIX 1+5
#define NINE 8+1

int main(void)
{
printf("%d * %d = %d\n", SIX, NINE, SIX * NINE);
return 0;
}

--
Keith Thompson (The_Other_Keit h) kst-u@mib.org <http://www.ghoti.net/~kst>
San Diego Supercomputer Center <* <http://users.sdsc.edu/~kst>
We must do something. This is something. Therefore, we must do this.