NetNews Offline 2/NetNews Offline Volume 2.iso/news/comp/lang/c-part1/7136

Internet Message Format | 1996-08-05 | 2.0 KB

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From: Lawrence Kirby <[e-mail]>
Newsgroups: comp.lang.c
Subject: Re: calloc help needed
Date: Fri, 16 Feb 96 20:47:41 GMT
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Message-ID: <[e-mail]>
References: <[e-mail]> <[e-mail]>
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In article <[e-mail]>
           [e-mail] "Kazimir Kylheku" writes:

>In article <[e-mail]>,
>A M Casey <[e-mail]> wrote:
> >I'm trying to allocate enough memory for an array of strings, defined as
> >char * MyArray[20].

That defines an array of 20 pointers. You will have to set each pointer
individually to point into a suitable array of characters.

> >the only manual entry I have for calloc says that it needs two values, the
> >number of elements, and the size of each, but how do I actually use it?
> >
> >MyArray = calloc(20,50);

This is illegal with the definition of MyArray you have (you can't assign
to an array). What you may want is:

   char (*MyArray)[50];

where I take 50 to be the wize of each character array to be allocated for
each string. However see the FAQ for more information about allocating
'2D' arrays.

>This is gives you much more space than you need for the array. A character's
>size is not 50!

But he asked for an array of 'strings' not an array of characters.

> >doesnt work, as far as I can tell calloc returns an int.

ANSI calloc returns void *.

>It is probably "implicitly defined" to return an int, because you lack a
>declaration. Try including the "malloc.h" header file.

The correct header file to include is stdlib.h.

--
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Lawrence Kirby | [e-mail]
Wilts, England | [e-mail]
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