Path: utzoo!utgpu!jarvis.csri.toronto.edu!mailrus!cornell!uw-beaver!ssc-vax!dmg
From: dmg@ssc-vax.UUCP (David Geary)
Newsgroups: comp.sys.amiga
Subject: Re: Struggling through the "C"
Message-ID: <2606@ssc-vax.UUCP>
Date: 14 Apr 89 22:19:37 GMT
Organization: Boeing Aerospace Corp., Seattle WA
Lines: 99


In article <231@nlgvax.UUCP>, Hans Zuidam writes:

||In article <97113@sun.Eng.Sun.COM>, Chuck McManis writes:
|||char	*a; /* This is a pointer to a string */
|||char	a[80]; /* this is a pointer to a string with 80 bytes allocated */
|
|In article <2570@ssc-vax.UUCP> dmg@ssc-vax.UUCP (David Geary) writes:
||  No!  char a[80] is not a pointer at all.  It is an array...
|
|Please check your K&R before responding:
|
|	K&R 7.1, pg. 185:
|    An identifier is a primary expression, provided it has been
|    suitably declared as discussed below. If the type of the identifier
|    is "array of ...", then the value of the identifier-expression is a
                                 ^^^^^^^^

  Please *understand* your K&R before responding ;-)

  Of course, this is true - the *value* of the identifier-expression is a
pointer to the first object in the array.  What this means is:

main()
{
  char a[80];	/* an ARRAY - NOT a pointer */

  DoSomething(a);  /* here, the value of a is equivalent to &a[0], which
		      is a pointer to the first object in the array a.
		      This means we could have written:
		      DoSomething(&a[0]); and gotten the same result. */
}

|    pointer to the first object in the array, and the type of the
|    expression is "pointer to ...".
|
|and
|
|    K&R 7.1, pg. 186:
|    The expression E1[E2] is identical (by definition) to *((E1)+(E2)).

  Yes, this is true.

|
|So:
|    'char *a' is *identical* to 'char a[80]' *except* for the
|    fact that in the second definition 'a' is a *constant*
|    pointer. That is: you cannot assign a value to 'a'.

  No! (no no no ;-) - This is a common misconception that many of my
students in my introductory C class make.  char *a is NOT identical
to char a[80];  While it's true that in char a[80], the *value*
of a is a constant pointer, and in char *a, a is a variable pointer, 
there is more difference than that:

char *a;

  This gives you a pointer, period.  a can hold the address where
a char sits in memory:

---------
|       |----------->
---------

char a[80];

  This gives you an actual array of characters, there is no pointer
involved:

---------
| a[0]  |
---------
| a[1]  |
---------
| a[2]  |
---------
   .
   .
   .

  If you think that char *a and char a[80] are *identical* , *except*
for the fact that the second definition...(as stated above), try this:

main()
{
  char a[80], *p;

  strcpy(a, "I can fit in array a, because there is memory allocated for me");
  strcpy(p, "This is nothing but trouble");
}

Notice that the second strcpy is going to cause you severe headaches
(especially on an Amiga).

-- 
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
~ David Geary, Boeing Aerospace, Seattle                 ~ 
~ "I wish I lived where it *only* rains 364 days a year" ~
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~