Path: utzoo!attcan!uunet!lll-winken!lll-lcc!ames!ncar!gatech!fabscal!mpx1!vu-vlsi!devon!shockeye!hermit
From: hermit@shockeye.UUCP (Mark Buda)
Newsgroups: comp.lang.c
Subject: Re: pointer allocation
Message-ID: <275@shockeye.UUCP>
Date: 20 Dec 88 19:09:44 GMT
References: <8Xf3eSy00UkZ40cVE0@andrew.cmu.edu>
Reply-To: hermit@shockeye.UUCP (Mark Buda)
Organization: The Lab Rats
Lines: 36

In article <8Xf3eSy00UkZ40cVE0@andrew.cmu.edu> xj01+@andrew.cmu.edu (Xue Jue) writes:
>I am try to allocate a list of integer pointers by the following short program.
>But for some reason, after I compiled it and tried to run, it gave me a
>Segmentation fault message. I couldn't see what's wrong there. Could anyone out
>there give me some guide?
>
>
>#include <stdio.h>
>int *(*B);
>
>main()
>{
> *B = (int *) calloc(18,sizeof(int));
>}

Wow, a question I can answer, if not clearly (or correctly) :-)

Problem 1: You are allocating memory for 18 ints, not 18 pointers to ints.
	Try calloc(18,sizeof(int *));
Problem 2: If you were allocating memory for pointers to ints, you would
	want to cast the value returned by calloc to an (int **), not an (int *).
Problem 3: You are storing the pointer returned by calloc() in *B. This is
	very very bad. What you want is to store it in B. Storing it in *B puts
	it in the memory location B points to, and since B is never given a
	value, it doesn't really point anywhere useful. In fact, it is probably
	NULL. And as we all know, dereferencing a NULL pointer can mess up your
	day.

>By the way, this is on IBM RT work station. The same thing has been running on
>IBM PC with microsoft C 5.0.

Shows you something about the IBM PC... :-)
-- 
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