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From: pjh@mccc.edu (Pete Holsberg)
Newsgroups: comp.lang.c
Subject: Re: Strange Behavior?
Message-ID: <1991May9.161425.17079@mccc.edu>
Date: 9 May 91 16:14:25 GMT
References: <1991May8.020720.20170@mccc.edu> <1991May8.204009.1694@cubmol.bio.columbia.edu>
Organization: The College On The Other Side Of Route One
Lines: 37

In article <1991May8.204009.1694@cubmol.bio.columbia.edu> ping@cubmol.bio.columbia.edu (Shiping Zhang) writes:
=In article <1991May8.020720.20170@mccc.edu> pjh@mccc.edu (Peter J. Holsberg) writes:
=>Here is an extract from a program a student wrote.  Note that he has
=>declared "input" incorrectly.  The strange behavior is that, when choice
=>"1" is made, the print function outputs all but the first line that was
=>entered.  Can anyone explain that in terms of what scanf() is doing to
=>memory near "input"?  (This is on a 386, if endianism matters.)
= 
=[...]
=
=>   char sentence [MAX][SIZE];   /* an array of strings */
=>   char *point[MAX];            /* an array of pointers to char */
=>   char *orginal[MAX];          /* an array holding the orginal sequence */
=>   char input;	/* SHOULD HAVE BEEN int input!!!  */
= 
=[...]
= 
=>	printf("Make a choice: ");
=>	scanf ("%d" , &input);      /* value converted to decimal integer
=
=I think it results from declaring input as char and using it as int
=in scanf(). When scanf() writes 1 into the location of input, it puts
=into 0's those bits that belong to the first byte(s) of sentence[0],
=terminating sentence[0] at its first byte.  I could be wrong, though.

I'm sure you're right.  But what I really want to know is why the
compiler assigns memory that way.  It would seem to me that it would
assign memory "downward," so that the attempt to write a 4 byte value
into a 1 byte location would result in the "corruption" of the 3 cells
*following* the one allocated to "input."

Pete
-- 
Prof. Peter J. Holsberg      Mercer County Community College
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