Path: utzoo!attcan!uunet!cs.utexas.edu!jarvis.csri.toronto.edu!torsqnt!hybrid!robohack!druid!darcy
From: darcy@druid.uucp (D'Arcy J.M. Cain)
Newsgroups: comp.lang.c
Subject: Re: FREE
Message-ID: <1990Feb27.155133.20341@druid.uucp>
Date: 27 Feb 90 15:51:33 GMT
References: <2714@stl.stc.co.uk>
Reply-To: darcy@druid.UUCP (D'Arcy J.M. Cain)
Organization: D'Arcy Cain Consulting, West Hill, Ontario
Lines: 42

In article <2714@stl.stc.co.uk> dsr@stl.stc.co.uk (David Riches) writes:
>We're writing a tool which requires memory management.
>We have a routine which allocates memory in the following
>fashion :-
>  #define NE_ARR_MALLOC(y,n) ((y *) malloc ((unsigned) ((n) * (sizeof(y))))) 
>
>Which will allocate space for n of y.
calloc does what you want and even initialize the space to zeroes for you.
To allocate n of y use:
    x = (y *)calloc(n, sizeof(y);
where x is a pointer to type y.  If using an ANSI compiler you can even leave
off the cast since calloc (and malloc) returns a pointer to void.

>
>The question then arises as how best to free this space?
>In most cases only the pointer to the space is known but will the
>following free up all the space :-
>  #define NE_ARR_FREE(x)     { free((char *) sizeof(x)); x = 0; }
>
>What happens to other references which might be pointing half way
>down the list for instance? 
>
I'm afraid I can't parse the above statement.  It seems to be saying that
the argument to free is the size of the array casted to a pointer.  In any
case to free up space simply pass free the pointer to the malloc'ed space.

    struct y *x;
	unsigned nelems = NUMBER_OF ELEMENTS_REQUIRED_FOR_YOUR_APPLICATION;

    ...
    x = malloc(nelems * sizeof(y));
        /***** OR *****/
    x = calloc(nelems, sizeof(y));
    ...
    free(x);


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D'Arcy J.M. Cain (darcy@druid)     |   Thank goodness we don't get all 
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