Hi,
When you define an array base with a define statement, like
#define ARRAY_BASE 0x1000;
and access it, such as
ARRAY_BASE[x]
how does the compiler distinguish the offset? It is one-word wide by
default perhaps? (i.e. usually 32 bits)
Bahadir 9 2129 ba************@ gmail.com wrote: When you define an array base with a define statement, like
#define ARRAY_BASE 0x1000;
and access it, such as
ARRAY_BASE[x]
You don't; that is a syntax error. One of the operands to [] must be a
pointer, which isn't the case here unless x is one.
Richard
In article <11************ *********@z14g2 000cwz.googlegr oups.com>,
<ba************ @gmail.com> wrote: #define ARRAY_BASE 0x1000;
Presumably you didn't mean the semicolon there.
ARRAY_BASE[x]
I'm assuming you intend x to be an integer.
how does the compiler distinguish the offset?
It doesn't. It prints out an error message, such as "subscripte d
value is neither array nor pointer".
-- Richard
Richard Tobin wrote: In article <11************ *********@z14g2 000cwz.googlegr oups.com>, <ba************ @gmail.com> wrote:
#define ARRAY_BASE 0x1000;
Presumably you didn't mean the semicolon there.
ARRAY_BASE[x]
I'm assuming you intend x to be an integer.
how does the compiler distinguish the offset?
It doesn't. It prints out an error message, such as "subscripte d value is neither array nor pointer".
-- Richard
Perhaps the former was my lamest post in last 5 years. Sorry for that.
The place I saw it was actually with an int * cast, and obviously that
cast determines the offset. I just got confused. Sometimes your mind
just stops you know.
Bahadir
Emmanuel Delahaye wrote: ba************@ gmail.com wrote on 18/03/05 :
#define ARRAY_BASE 0x1000;
and access it, such as
ARRAY_BASE[x]
This expands to
0x1000;[x]
obvioulsy, it's not C...
Note that
#define ARRAY_BASE 0x1000
ARRAY_BASE[x];
is not C either because an integer and a pointer are different animals.
But this does -
#define ARRAY_BASE 1000
and access it, such as
ARRAY_BASE[x] = <value>;
perhaps a typo in O.P code !
- Ravi
Ravi Uday wrote: Emmanuel Delahaye wrote:
ba************@ gmail.com wrote on 18/03/05 :
#define ARRAY_BASE 0x1000;
and access it, such as
ARRAY_BASE[x]
This expands to
0x1000;[x]
obvioulsy, it's not C...
Note that
#define ARRAY_BASE 0x1000
ARRAY_BASE[x];
is not C either because an integer and a pointer are different animals. But this does -
Does what?
There's no previous "doesn't" in this thread for reference.
#define ARRAY_BASE 1000
and access it, such as
ARRAY_BASE[x] = <value>;
perhaps a typo in O.P code !
Is x supposed to be a pointer?
--
pete
Ravi Uday wrote: Emmanuel Delahaye wrote:
<snip> Note that
#define ARRAY_BASE 0x1000
ARRAY_BASE[x];
is not C either because an integer and a pointer are different animals.
^^^^^^^^^^^^^^^ ^^^^^^^^^^^^^^^ ^^^^^^^^^^^^^^^ But this does -
#define ARRAY_BASE 1000
and access it, such as
ARRAY_BASE[x] = <value>;
perhaps a typo in O.P code !
No, that is invalid for the reason Emmanuel Delahaye stated above.
Integers and pointers are not the same thing. All you have done is
change the value of the integer from 1000 hex to 1000 decimal probably
making it even worse.
The "correct" code if your system has something at address 1000 hex that
you are both allowed to access as an array of some_type and want to so
access would be
#define ARRAY_BASE ((some_type *)0x1000)
ARRAY_BASE[x];
This invokes undefined behaviour, however things like this are not too
uncommon in embedded systems where you have a memory mapped hardware
device you want to access.
--
Flash Gordon
Living in interesting times.
Although my email address says spam, it is real and I read it.
Flash Gordon wrote: ... But this does -
#define ARRAY_BASE 1000
and access it, such as
ARRAY_BASE[x] = <value>;
perhaps a typo in O.P code !
No, that is invalid for the reason Emmanuel Delahaye stated above. Integers and pointers are not the same thing. All you have done is change the value of the integer from 1000 hex to 1000 decimal probably making it even worse.
The important thing is that ';' is removed. In this case
ARRAY_BASE[x] = <value>;
has a chance to become a valid statement. Nothing says that 'x' must be
an integer. 'x' could be a pointer or an array, in which case the above
statement is perfectly valid.
#define ARRAY_BASE 1000
int x[2000];
ARRAY_BASE[x] = 5;
--
Best regards,
Andrey Tarasevich ba************@ gmail.com writes:
[...] Perhaps the former was my lamest post in last 5 years. Sorry for that. The place I saw it was actually with an int * cast, and obviously that cast determines the offset. I just got confused. Sometimes your mind just stops you know.
It happens to all of us now and then.
--
Keith Thompson (The_Other_Keit h) ks***@mib.org <http://www.ghoti.net/~kst>
San Diego Supercomputer Center <*> <http://users.sdsc.edu/~kst>
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