Re: [9fans] thread

Re: [9fans] thread

[dmr]

So far as I can determine, the Plan 9 C compiler is
conformant here.

If you have, say, an

	int A[10];

then just A is the address of A and has type int *.
&A is a pointer to this array of 10 integers; it will
have the same value, as a pointer, but a different type:

 int (*)[10]

The reason why the initialization in the thread man page,
(and presumably the code) works is that the thing being
assigned to is void *, and any object pointer will fit; the
coercion just happens.

The & in &m is redundant, probably misleading, but ultimately
harmless.  Kenji's example with print likewise has the
same character, since print is a variadic function, and all
his second arguments fall under the ... in its declaration,
and so aren't type-checked.

Incidentally, there is no C-language guarantee that the
actual bit pattern produced by A and &A will be identical,
though in practice they usually will be for most compilers,
yet they still have different types.

For a case in which the type-checking is more evident, without
the laxness of void * or ..., try compiling

#include <u.h>
#include <libc.h>

void t1(int *ip);
void t2(int (*iap)[10]);

void
main(void)
{
	int ia[10];

	t1(ia);
	t1(&ia);
	t2(ia);
	t2(&ia);
}

You will get errors on lines 13 [ t1(&ia) ] and
14 [ t2(ia) ].

	Dennis

Re: [9fans] thread

[Digby Tarvin]

Hi,

That is a fascinating little subtlety which I hadn't appreciated
before.

I was aware that the array type was special in that the application of
the '&' operator to it resulted in a change of type but not of value.

But if I understand correctly, there is a converse situation with
the type '(*)[n]' (ie address of an array) where the application
of the unary '*' operator also produces a change of type but not value.

Or said another way, the address of an array of ints is a different type
to an array of ints, but for any given array the two values are identical:

main()
{
        int     a[1];
        int     (*la)[1];

        a[0] = 10;
        la = &a;
        printf("  a = %x\n", a);	/* array of ints */
        printf(" &a = %x\n", &a);	/* address of array of ints */
        printf(" la = %x\n", la);	/* address of array of ints */
        printf("*la = %x\n", *la);	/* array of ints */
}

Which produces (gcc on BSD - sorry, Plan9 is not at hand)
	  a = efbfd91c
	 &a = efbfd91c
	 la = efbfd91c
	*la = efbfd91c

Subtle, but at least it is symmetric.

Regards,
DigbyT

dmr@plan9.bell-labs.com:
> So far as I can determine, the Plan 9 C compiler is
> conformant here.
>
> If you have, say, an
>
> 	int A[10];
>
> then just A is the address of A and has type int *.
> &A is a pointer to this array of 10 integers; it will
> have the same value, as a pointer, but a different type:
>
>  int (*)[10]
>
	[snip]
>
> 	Dennis
--
Digby R. S. Tarvin                                              digbyt@acm.org
http://www.cthulhu.dircon.co.uk

Re: [9fans] thread

[Douglas A. Gwyn]

Digby Tarvin wrote:
> I was aware that the array type was special in that the application of
> the '&' operator to it resulted in a change of type but not of value.

? One cannot (meaningfully) apply & to a type, only to a certain kind
of expression.  When that expression consists of an identifier that
has been declared as an object having array type, then in fact
&that_identifier is *not* treated specially, according to the C
standard, but rather follows the generic rules for the & operator.
What is special is the special rule that the unadorned identifier
decays, in most contexts, into a pointer to the first element of the
array designated by that identifier; this is closely connected with
arrays not being first-class objects in C.  It is often convenient,
but is not natural.

Re: [9fans] thread

[Digby Tarvin]

I think you misunderstand. I was referring to the application of '&' to
a variable of array type, not to a type. Applied to an array type, it
produces a change of type but no change in value. An unusual case
where the '&' operator could be replaced by a cast with no change
in semantics.  I would say that was special treatment, (resulting
from the special treatments of array names) albeit in accordance
with the C standard.

Regards,
DigbyT

Douglas A. Gwyn:

> Digby Tarvin wrote:
> > I was aware that the array type was special in that the application of
> > the '&' operator to it resulted in a change of type but not of value.
>
> ? One cannot (meaningfully) apply & to a type, only to a certain kind
> of expression.  When that expression consists of an identifier that
> has been declared as an object having array type, then in fact
> &that_identifier is *not* treated specially, according to the C
> standard, but rather follows the generic rules for the & operator.
> What is special is the special rule that the unadorned identifier
> decays, in most contexts, into a pointer to the first element of the
> array designated by that identifier; this is closely connected with
> arrays not being first-class objects in C.  It is often convenient,
> but is not natural.
>
--
Digby R. S. Tarvin                                              digbyt@acm.org
http://www.cthulhu.dircon.co.uk

Re: [9fans] thread

[Boyd Roberts]

From: "Douglas A. Gwyn" <DAGwyn@null.net>
> Digby Tarvin wrote:
> > I was aware that the array type was special in that the application of
> > the '&' operator to it resulted in a change of type but not of value.
>
> ? One cannot (meaningfully) apply & to a type, only to a certain kind
> of expression.  When that expression consists of an identifier that
> has been declared as an object having array type, then in fact
> &that_identifier is *not* treated specially, according to the C
> standard, but rather follows the generic rules for the & operator.

the one time illegal &array construct falls out of the history of C.
the reason being this is that an array identifier was not an lvalue,
but an lvalue-expression.

read what dennis has to say at:

    http://cm.bell-labs.com/cm/cs/who/dmr/chist.html

ps. i know doug knows this.

pps. there's http://cm.bell-labs.com/cm/cs/who/dmr/cman.pdf too.

Re: [9fans] thread

[Dan Cross]

In article <200107091146.MAA27082@cthulhu.dircon.co.uk> you write:
>I think you misunderstand. I was referring to the application of '&' to
>a variable of array type, not to a type. Applied to an array type, it
                                                     ^^ a variable of
>produces a change of type but no change in value.

Errm, isn't this implementation dependent?

>An unusual case
>where the '&' operator could be replaced by a cast with no change
>in semantics.  I would say that was special treatment, (resulting
>from the special treatments of array names) albeit in accordance
>with the C standard.

But the semantics are different, as Dennis pointed out in his post.

	- Dan C.

Re: [9fans] thread

[Boyd Roberts]

> However, Dennis clarified how it ...

he did, but do not forget that he said that printf
is varadic.

do not forget either that using %d with pointers (to int
or anything thing else) is fraught with peril.

Re: [9fans] thread

[Dan Cross]

In article <20010706042502.AE409199D7@mail.cse.psu.edu> you write:
>Thank you, Dan,

Thanks, Kenji, though I think that Dennis' note was better.  :-)

#include <u.h>
>#include <libc.h>
>main(){
>	int a[1];
>	a[0] = 10;
>	print("%d\n",*a);
>	print("%d\n",a);
>	print("%d\n",&a);
>	print("%d\n",*&a);
>	print("%d\n",*(&a));

These are the two that confused me, since they output:

>2147479448
>2147479448

However, Dennis clarified how it worked when he pointed out the
type differences between a and &a (the former is int *, the latter
int (*)[1]).  I believe that in the case of *&a, you end up with
a variable of type a[1], which decays into int *.  Restated another
way, it's as if the * and & cancel each other out, leaving you
an unadorned a, which turns into int *.

At least, I believe that's how it works; I'd love to be corrected
if I'm wrong.  Since this isn't comp.lang.c, I can actually look
forward to such a thing.  ;-p

	- Dan C.

Re: [9fans] thread

[arisawa]

Thank you, Dan,

You say:
>No, &m in this context is perfectly legal C.  ``&'' in C says
>to take the address of the referenced object (m)

But, how can I understand the following result?

#include <libc.h>
main(){
	int a[1];
	a[0] = 10;
	print("%d\n",*a);
	print("%d\n",a);
	print("%d\n",&a);
	print("%d\n",*&a);
	print("%d\n",*(&a));
	print("%d\n",*(a));
}

The output is:
10
2147479448
2147479448
2147479448
2147479448
10

Kenji Arisawa
E-mail: arisawa@aichi-u.ac.jp

Re: [9fans] thread

[Dan Cross]

In article <20010706010009.D94AF199C0@mail.cse.psu.edu> you write:
>Hello 9fans,

Hey Kenji,

>Manual of thread shows the example codes such as:
>	char m[48];
>	...
>	{nil,   &m,     CHANRCV},
>I wonder that &m is a typo.

No, &m in this context is perfectly legal C.  ``&'' in C says
to take the address of the referenced object (m), where m refers
to that 48 char array.  However, ``m'' by itself will also decay
into a pointer to the array, so the two uses are equivalent in
this context.  At least, that's the gist of it; Doug Gywn would
be able to describe it more precisely than I have.

As Rob says, though, the & is irrelevant, and I've seen it be
confusing in the past; particularly for junior programmers.  I
recommend writing C code which references arrays without the &.

	- Dan C.

Re: [9fans] thread

[rob pike]

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Unnecessary, anyway.

-rob


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From: arisawa@ar.aichi-u.ac.jp
To: 9fans@cse.psu.edu
Subject: [9fans] thread
Date: Fri, 6 Jul 2001 09:50:20 +0900
Message-ID: <20010706010009.D94AF199C0@mail.cse.psu.edu>

Hello 9fans,

Manual of thread shows the example codes such as:
	char m[48];
	...
	{nil,   &m,     CHANRCV},
I wonder that &m is a typo.

Thanks,

Kenji Arisawa
E-mail: arisawa@aichi-u.ac.jp

[9fans] thread

[arisawa]

Hello 9fans,

Manual of thread shows the example codes such as:
	char m[48];
	...
	{nil,   &m,     CHANRCV},
I wonder that &m is a typo.

Thanks,

Kenji Arisawa
E-mail: arisawa@aichi-u.ac.jp