NetNews Offline 2/NetNews Offline Volume 2.iso/news/comp/lang/c-part1/9327

Internet Message Format | 1996-08-05 | 2.3 KB

Path: hubcap.clemson.edu!hubcap!mjs
From: [e-mail] (M. J. Saltzman)
Newsgroups: comp.lang.c
Subject: Re: NEwbie: How to return a multi-dimensional array from function?
Date: 8 Mar 96 15:23:49 GMT
Organization: Clemson University
Message-ID: <mjs.826298629@hubcap>
References: <4hp273$[e-mail]> <[e-mail]>
NNTP-Posting-Host: hubcap.clemson.edu
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Sean Park <[e-mail]> writes:

>Anthony Moendir wrote:
>>
>> I want to return a multidimensional array fro a function,
>> something like this:
>>
>> char *Foo();
>>
>> int main()
>> {
>>  char tmp[10][5];
>>
>>  tmp=Foo();
>> }
>>
>> char *Foo()
>> {
>>  char tmp[10][5];

Two things: (1) If you are going to use the value of tmp outside of Foo(),
then you need to declare

    static tmp[10][5];

Otherwise the storage may disappear when you return from Foo().

>> //do something
>> return(tmp);
         ^   ^ Parentheses not required here.
>> }

(2) When you use tmp in an expression, the compiler replaces the array
name with a constant pointer to its first element.  (Among other
consequences, this means that arrays can't appear on the left side of
assignments.)  In this case, the first element has type
array-of-15-chars, so a pointer to it is a
pointer-to-array-of-15-chars.  That's the type that the function
should return, and the type of the variable that you assign it to.
Thus, Foo() needs to be declared

    char (*Foo())[5];

and the tmp in main() should be

    char (*tmp)[5];

>> How should i do that?
>>
>> Any help would be appreciated
>> Anthony

Another technique for getting the result back is to pass the array in as an
argument, as is done below (with corrections).

>You could pass the pointer to the array to the function.  That way the
>function itself will fill your array addresses rather than returning
>redundant data.  You might try:

>void foo(char *tmp);
          ^^^^^^^^^ Here, tmp should be declared char (*tmp)[5].
>int main()
>{
>   char tmp[10][5];
>   foo(tmp);
>   return 0;
>}
>void foo(char *tmp)
          ^^^^^^^^^ Here, tmp should be declared char (*tmp)[5].
>{
>/* You can reference tmp within this function as an array of what ever
> * dimensions you want. */
>/* do something */

> return;
>}

>HTH

>Sean Park
>[e-mail]
--
        Matthew Saltzman
        Clemson University Math Sciences
        [e-mail]