NetNews Offline 2/NetNews Offline Volume 2.iso/news/comp/lang/c-part1/3200

Text File | 1996-08-05 | 1.7 KB | 83 lines

Path: atglab.bls.com!Alun.Champion
From: [e-mail] (Alun Champion)
Newsgroups: comp.lang.c
Subject: Re: Can a function return a (pointer to a) function?
Date: 26 Jan 1996 23:14:42 GMT
Organization: Computer People Inc.
Message-ID: <[e-mail]>
References: <4ebaaa$[e-mail]>
NNTP-Posting-Host: bstfirewall.bst.bls.com
In-reply-to: "John D. Hourihane"'s message of 26 Jan 1996 19:33:30 GMT

In article <4ebaaa$[e-mail]> "John D. Hourihane" <[e-mail]> writes:

: I am trying to write a function which will return a pointer to a function,
: but I'm having no luck.

Your problem lies in the declaration of the function.
  (int f(int, int)) *pick(int s);
It's a very logical attempt but unfortunately not quite right.
It should have been:
  int (*pick(int s))(int,int);

I would look on the net for a utility called cdecl, it will explain these
complication declarations for you.

Typedefs are very good for this sort of thing and makes things a lot
simpler to understand:

  typedef int (*FuncPtr)(int, int);
  FuncPtr pick(int s);

Here's your code corrected:

#include <stdio.h>

#define ADD 0
#define MULTIPLY 1

typedef int (*FuncPtr)(int, int);

FuncPtr pick(int s);

int
main(void)
{
    printf("5 + 4 = %d\n", pick(ADD)(5,4));
    printf("5 * 4 = %d\n", pick(MULTIPLY)(5,4));

    return 0;
}

int
add(int x, int y)
{
    return x + y;
}

int
multiply(int x, int y)
{
    return x * y;
}

FuncPtr pick(int s)
{
    switch (s) {
    case ADD:
        return add;
    case MULITPLY:
        return multiply;
    }

    return 0;
}

Hope this helps

Regards

   -A.
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