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Array at specific address

Don
Hi I have an array of unsigned chars, like:
MyArray[2048] = {0x00};

For memory-mapping purposes I need to store this array at a specific address
(0xFFFF1199)

How do I declare this?

I cant't do MyArray[2048] = {0x00} @ 0xFFFF1199 :-(

I am using the gnu compiler.

Best Regards
Don
Nov 13 '05
24 16340
On 25 Sep 2003 15:33:28 -0700, Micah Cowan <mi***@cowan.na me> wrote in
comp.lang.c:
Joona I Palaste <pa*****@cc.hel sinki.fi> writes:
Don <no**@spam.dk > scribbled the following:
Hi I have an array of unsigned chars, like:
MyArray[2048] = {0x00};

For memory-mapping purposes I need to store this array at a specific address
(0xFFFF1199)

How do I declare this?


You can't. Not under ISO standard C. On some implementations , you can't
declare it at all.


Of course you can.

memcpy((void*)0 xFFFF1199, MyArray, sizeof MyArray);

Now, whether or not this does what you want is
implementation-defined, but you certainly can do it, if it's
doable. The Standard specifically says so.

-Micah


Exactly how does this declare an array of 2048 unsigned characters at
all?

BTW, you're another one who could use a proper signature separator.

--
Jack Klein
Home: http://JK-Technology.Com
FAQs for
comp.lang.c http://www.eskimo.com/~scs/C-faq/top.html
comp.lang.c++ http://www.parashift.com/c++-faq-lite/
alt.comp.lang.l earn.c-c++ ftp://snurse-l.org/pub/acllc-c++/faq
Nov 13 '05 #11
Hi,

To declare an array in C:
{
char temp[100] ;

}

Compiler must know its size. If the array is declared in a context the
compiler places it in stack and if a global it places it in heap.

Now how to place an array at a specified location.

I think a static array u can't place it.
With a pointer u can do it.

char * p = (char *) 1000 ;
Now u can use p like u use an array.

p [10] works

Hope this helps,

Regards,
James
Nov 13 '05 #12
Mark McIntyre <ma**********@s pamcop.net> wrote:
On Thu, 25 Sep 2003 10:02:07 GMT, in comp.lang.c , Kevin Easton
<kevin@-nospam-pcug.org.au> wrote:
Derk Gwen <de******@hotpo p.com> wrote:
"Don" <no**@spam.dk > wrote:
# Hi I have an array of unsigned chars, like:
# MyArray[2048] = {0x00};

If it _has_ to be an array instead of (unsigned char*), many linkers allow you to
specify the address of extern objects. You might try checking what your linker
permits.


You can do it without linker magic also:

#define MyArray (*(volatile unsigned char (*)[2048])(0xdeadbeef))


When I tried this on my DS9K, it said "moo" and printed out a recipe
for making cow from hamburger.


A legal preprocessor directive did all that? I think you need to get a
conforming implementation (I'm reliably informed that one exists for the
DS9000).

- Kevin.

Nov 13 '05 #13
On Fri, 26 Sep 2003 00:00:56 +0200, in comp.lang.c , "Martijn"
<su************ *********@hotNO FILTERmail.com> wrote:
#define MyArray (*(volatile unsigned char (*)[2048])(0xdeadbeef))


When I tried this on my DS9K, it said "moo" and printed out a recipe
for making cow from hamburger.


liar


No, thats a greek musical instrument, beloved by muses the world over.
I do admit its often strung with cowgut though.

--
Mark McIntyre
CLC FAQ <http://www.eskimo.com/~scs/C-faq/top.html>
CLC readme: <http://www.angelfire.c om/ms3/bchambless0/welcome_to_clc. html>
Nov 13 '05 #14
On Fri, 26 Sep 2003 10:32:04 GMT, in comp.lang.c , Kevin Easton
<kevin@-nospam-pcug.org.au> wrote:
Mark McIntyre <ma**********@s pamcop.net> wrote:
On Thu, 25 Sep 2003 10:02:07 GMT, in comp.lang.c , Kevin Easton

#define MyArray (*(volatile unsigned char (*)[2048])(0xdeadbeef))
When I tried this on my DS9K, it said "moo" and printed out a recipe
for making cow from hamburger.


A legal preprocessor directive did all that?


The message appeared at runtime. The result of the directive is just a
tad implementation defined, I believe.
I think you need to get a
conforming implementation (I'm reliably informed that one exists for the
DS9000).


The one I have is as conforming as it gets, if somewhat
pathological...
--
Mark McIntyre
CLC FAQ <http://www.eskimo.com/~scs/C-faq/top.html>
CLC readme: <http://www.angelfire.c om/ms3/bchambless0/welcome_to_clc. html>
Nov 13 '05 #15
Jack Klein <ja*******@spam cop.net> writes:
On 25 Sep 2003 15:33:28 -0700, Micah Cowan <mi***@cowan.na me> wrote in
comp.lang.c:
Joona I Palaste <pa*****@cc.hel sinki.fi> writes:
Don <no**@spam.dk > scribbled the following:
> Hi I have an array of unsigned chars, like:
> MyArray[2048] = {0x00};

> For memory-mapping purposes I need to store this array at a specific address
> (0xFFFF1199)

> How do I declare this?

You can't. Not under ISO standard C. On some implementations , you can't
declare it at all.
Of course you can.

memcpy((void*)0 xFFFF1199, MyArray, sizeof MyArray);

Now, whether or not this does what you want is
implementation-defined, but you certainly can do it, if it's
doable. The Standard specifically says so.

-Micah


Exactly how does this declare an array of 2048 unsigned characters at
all?


I never said it did. It does, however, answer the question "I
need to store this array at a specific address." He could use
memset() to better effect, though. And, of course, if he wants to
access it literally as an array, he can always do:

unsigned char (*MyArray)[2048] = (unsigned char (*)[2048])0xFFFF1199;

and dereference MyArray.

BTW, you're another one who could use a proper signature separator.


Huh. I'd always been under the impression that it was mainly for
those with big long sigs.

--
Micah
(this better?)
Nov 13 '05 #16
ja***********@y ahoo.co.uk (James) writes:
Hi,

To declare an array in C:
{
char temp[100] ;

}

Compiler must know its size. If the array is declared in a context the
compiler places it in stack and if a global it places it in heap.
That's not mandated by the standard, so it's
implementation-specific information (even if most existing
implementations happen to implement that way).
Now how to place an array at a specified location.

I think a static array u can't place it.
You can't place any array, regardless of whether static, dynamic
or automatic allocation is used. If you can write to the memory
at location "1000", you haven't allocated at all: it's simply there.
With a pointer u can do it.

char * p = (char *) 1000 ;
Now u can use p like u use an array.

p [10] works


No, you can't use p quite like an array. You use it more like a
pointer-to-char (which incidentally, is pretty much how you use
arrays most of the time). But among other things, you can't make
a call:

foo(&p)

to a function prototyped as:

void foo(char (*p)[1000]);

-Micah
Nov 13 '05 #17
Mark McIntyre <ma**********@s pamcop.net> wrote:
On Fri, 26 Sep 2003 10:32:04 GMT, in comp.lang.c , Kevin Easton
<kevin@-nospam-pcug.org.au> wrote:
Mark McIntyre <ma**********@s pamcop.net> wrote:
On Thu, 25 Sep 2003 10:02:07 GMT, in comp.lang.c , Kevin Easton

#define MyArray (*(volatile unsigned char (*)[2048])(0xdeadbeef))

When I tried this on my DS9K, it said "moo" and printed out a recipe
for making cow from hamburger.


A legal preprocessor directive did all that?


The message appeared at runtime. The result of the directive is just a
tad implementation defined, I believe.


No, the result of the directive is to introduce a macro called MyArray.
That's all.
I think you need to get a
conforming implementation (I'm reliably informed that one exists for the
DS9000).


The one I have is as conforming as it gets, if somewhat
pathological...


Not if merely defining a macro caused unrequested output at runtime.

- Kevin.

Nov 13 '05 #18
James wrote:
Now how to place an array at a specified location.

I think a static array u can't place it.
With a pointer u can do it.

char * p = (char *) 1000 ;
Now u can use p like u use an array.

p [10] works


char *p = (char *)10000; is legal C, but C offers absolutely no guarantees
that the address you've forced p to point to is a good place for storing
data. This is the kind of trick that it is best to avoid. For one thing,
it's considerably less portable than Brazil. For another thing, even if
it's legal to write to that space on your system (and it probably isn't, on
modern systems), where do you draw the line? p[100]? p[1000]?

(On the other hand, so-called "badly-behaved" MS-DOS programs used to do
this sort of thing all the time - for an excellent-at-the-time reason which
has become more or less irrelevant nowadays.)

--
Richard Heathfield : bi****@eton.pow ernet.co.uk
"Usenet is a strange place." - Dennis M Ritchie, 29 July 1999.
C FAQ: http://www.eskimo.com/~scs/C-faq/top.html
K&R answers, C books, etc: http://users.powernet.co.uk/eton
Nov 13 '05 #19
Richard Heathfield wrote:
char *p = (char *)10000; is legal C, but C offers absolutely
no guarantees that the address you've forced p to point to is
a good place for storing data. This is the kind of trick that
it is best to avoid. For one thing, it's considerably less
portable than Brazil. For another thing, even if it's legal to
write to that space on your system (and it probably isn't, on
modern systems), where do you draw the line? p[100]? p[1000]?

(On the other hand, so-called "badly-behaved" MS-DOS programs
used to do this sort of thing all the time - for an
excellent-at-the-time reason which has become more or less
irrelevant nowadays.)


It's not unusual to see this kind of code used in embedded
systems for which I/O and/or control spaces have been memory
mapped. I agree that it's not portable; but have found it
extremely useful within that limited context.
--
Morris Dovey
West Des Moines, Iowa USA
C links at http://www.iedu.com/c

Nov 13 '05 #20

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