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Internet Message Format | 1996-08-05 | 5.5 KB

Path: newshost.lanl.gov!tanmoy
From: [e-mail] (Tanmoy Bhattacharya)
Newsgroups: comp.lang.c
Subject: Re: What is &Variable (declared as: char Variable[10])?
Date: 27 Feb 1996 15:32:34 GMT
Organization: Los Alamos National Laboratory
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In-reply-to: Brenda's message of 26 Feb 1996 21:56:54 GMT

In article <4gtab6$[e-mail]>
Brenda <g051286> writes:
<snip>
B: >|>     char myarray[10];
<snip>
B: >Yes it could and yes it is...'myarray' is not a pointer, but &myarray is
B: >a pointer to 'myarray'.
B:
B: Um, that's not correct.  myarray is DEFINITELY a pointer!  As
declared above,
B: it is a constant pointer to 10 contiguous char datatypes.  myarray is an

Don't fill this newsgroup with garbage, we don't need this discussion
once again. Please chastise your service provider in the strongest
possible terms: it is common decency on their part not to flood proper
discussions with newbies who have no idea of nettiquette.

Most newsgroups have FAQs (Frequently Asked Questions with their
answers), and most are available by anonymous ftp from
rtfm.mit.edu. It is considered proper to retrieve them and read them
before posting, so as not to waste everybody's time with very common
misunderstandings. If you read and understand what the FAQ is saying,
and you think it is wrong, and you have enough reason to believe so,
that point is certainly welcome. Bringing it up without reading the
FAQ gets no one's sympathy.

In any case, it is polite to listen to a conversation a while before
jumping in. The same is true of USENET: and I am sure if you had
listened patiently for a week or two, you would have seen why you are
mistaken.

In any case, an array in C is not a pointer, just as `int i' does not
declare a pointer. `myarray' above and `i' in my declaration both
declare a region of storage, one for a sequence of 10 chars, and the
other for one int. `int *ptr', on the other hand, reserves storage for
a pointer to int. Note that `myarray' is not associated with any
storage for a pointer, and `ptr' is not associated with any storage
for an int. This can be easily verified by applying the `sizeof'
operator to each of them. One can easily verify as well that `char
myarray[10]' followed by `extern char *myarray' is a recipe for
disaster.

In most occurances of an object in an expression (for example when one
is not taking the address (&) or the sizeof the object), C evaluation
calls for the value of an object. The value of most objects is the
value stored in them, but the value of an array object is actually
independent of the value stored in the memory location that constitute
the array: it is instead a pointer to the _first_ of the sequence of
objects that constitute it. Thus it is completely incorrect to call it
a `constant pointer to 10 contiguous char datatypes'. What is true,
however, is that _unless_ it is the operand of & or sizeof (the actual
situation under discussion), the value of myarray is a `pointer to the
first of 10 contiguous char objects'. Note that it is very difficult
to talk of a value being a `constant', after i=5, is &i a constant
pointer to an int? Is i a constant integer 5?

B: ADDRESS whereas *(myarray + 5) or myarray[5] is the 6th element in
the array.

Yes. In *(myarray+5) or myarray[5] it is the value of myarray that
contributes. This indeed is a pointer.

B: The difference between an array and something like "char
*p=myarray", is that
B: you can say p++, but you can't say myarray++.  You shouldn't say &myarray

The ++ operator needs a `modifiable lvalue' which is a technical term.
What it means is that the expression must designate an object
(e.g. (i+j) will not do: that expression designates a value and not an
object), the object designator must not have a `const' qualified type,
or if it is a struct, union or array, none of its elements or fields
may have a const qualified type, (i.e. after `const int i', i++ is
disallowed), and it shouldn't be an array.

Actually, even if the C standard did not put in the special rule for
the array, you can easily see it wouldn't have made sense anyway. If
you wrote myarray = myarray + 1, by the above rules, myarray+1 would be
a pointer which you would have to assign to an array (if the array
hadn't yet decayed to a pointer) or to a value (if it had). Neither of
them makes the slightest sense, so, it is best not to even think of
these things.

B: either because myarray is a constant, but I read that on some compilers
B: scanf ignores the dereferencing and does not bother to warn you.

This is where you are most wrong. &myarray is very useful in some
contexts. It is a pointer to the entire array, and such pointers have
their uses. You are right in pointing out that passing them to scanf
invokes undefined behaviour.

B: Hope this helps.

Incorrect statements rarely help. Please read the FAQ, it is an
excellent source of information about C, and the product of a
considerable effort on the part of Steve Summit (and others).

Cheers
Tanmoy
--
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