Path: utzoo!attcan!uunet!cs.utexas.edu!tut.cis.ohio-state.edu!uccba!mead!dem
From: dem@mead.UUCP (Dave Myers)
Newsgroups: comp.lang.c
Subject: Re: What's a C expert?
Message-ID: <338@mead.UUCP>
Date: 24 Aug 89 14:06:49 GMT
References: <12214@well.UUCP> <6057@microsoft.UUCP> <1336@atanasoff.cs.iastate.edu>
Reply-To: dem@mead.UUCP (Dave Myers)
Distribution: all
Organization: Mead Data Central, Dayton OH.
Lines: 47


In article <1336@atanasoff.cs.iastate.edu> hascall@atanasoff.cs.iastate.edu.UUCP (John Hascall) writes:
>In article <6057> paulc@microsoft.UUCP (Paul Canniff 2/1011) writes:
>}In article <12214@well.UUCP> tmh@well.UUCP (Todd M. Hoff) writes:
> 
>}>           What do you need to know to be an expert C programmer?
> 
>}How about ... understands why a[i] equals i[a] and CAN EXPLAIN IT,
>
>    a[i] = *(a+i), i[a] = *(i+a), a+i = i+a
>


   Hmmm....  I don't claim to be a C expert, but let's take a trivial
case.  An array of anything with a size different from that of int
will do.

	int i;
	struct element {
		int x[BIGNUMBER];
	} a[WHATEVER];

	a[i] == *(a + (i * sizeof(element)))
	i[a] == *(i + (a * sizeof(int)))

	or

	a[i] == *(a + (i * sizeof(int) * BIGNUMBER))
	i[a] == *(i + (a * sizeof(int)))

   Clearly, these are not equal.  By the same (or at least a similar)
token, if you do,

	struct element *b;
	i = b;
	i++;
	b++;
	printf("%d", i == b);

you will get a 0, meaning that i and b were not incremented by the
same amount.

-- 
David Myers                                             (513) 865-1343   
Mead Data Central            This          Data Fabrication Technology
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