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From: hans@nlgvax.UUCP (Hans Zuidam)
Newsgroups: comp.sys.amiga
Subject: Re:  Struggling through the "C"
Message-ID: <231@nlgvax.UUCP>
Date: 7 Apr 89 11:44:06 GMT
References: <2570@ssc-vax.UUCP>
Reply-To: hans@nlgvax.UUCP (Hans Zuidam)
Followup-To: comp.sys.amiga
Organization: Philips Research Geldrop
Lines: 60

>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
    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)).
    
This is from:
    "The C programming language",
    Kernighan, Brian W., Ritchie, Dennis M.
    Prentice-Hall Software series, 1978.

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'.

Ponder these:
    c = "ABCDEF"[i]
    a[i] and i[a] and *(a + i)

    /* The function */
    f(a, b)
    char a[], b[];

    /* Called by */
    char *c, *d;
    f(c, d)

And so on.

A useful cosequence of this is that one can construct flexible
arrays:

     int *p, i;
     p = (int *) malloc(NrOfElements * sizeof(int));
     for (i = 0; i < NrOfElements; i++) p[i] =0;
     p = (int *) realloc(p, NewNrOfElements * sizeof(int));
     /* and so on... */

						Hans
-- 
Hans Zuidam                                    E-Mail: hans@pcg.philips.nl
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