Is this valid?
int a[20];
void *b;
b = (void *)a; // b points to a[0]
b += 5*sizeof(*a); // b points to a[5]
a[5] = 100;
printf( "%d\n" , *((int *)b) ); // prints 100
If so, if a had been a struct, would it still work?
Is there a possibility that the array could contain some padding, so
rather than sizeof, the assignment would be
b += 5*( (void *)(&(a[1])) - (void *)(&(a[0]));
which seems more more complex.
Would any padding be incorporated into sizeof anyway?
Sep 16 '08
160  5715  vi******@gmail. com wrote:
s0s...@gmail.co m wrote:
>vipps...@gmail .com wrote:
>>s0s...@gmail. com wrote:
[about incrementing void pointers as opposed to (char *) cast]
Because sometimes you don't need portability and because
sometimes there's a better way?
Yes, that better way being the portable way, in this case.
Well, it's a matter of whether you prefer void * or char *.
No you idiot, it's a matter of whether you prefer portability
over non-portability without any other gains or loses.
The point being that you can't increment a void* pointer in a
conformant C system. If you (sOs) look at the code generated you
will see that a cast to char* generates none, as does the cast back
to char*. However it clues the compiler in to what is going on.
--
[mail]: Chuck F (cbfalconer at maineline dot net)
[page]: <http://cbfalconer.home .att.net>
Try the download section. s0****@gmail.co m wrote:
>
.... snip ...
>
Shouldn't it be unsigned char *? Anyway, my argument was that if
your compiler can perform arithmetic with the void *, then go
void *, as you did above.
Because a compiler can't 'perform arithmetic with the void*'. It
says so, right in the C standard. Trying to do so results in
undefined behaviour.
--
[mail]: Chuck F (cbfalconer at maineline dot net)
[page]: <http://cbfalconer.home .att.net>
Try the download section.
CBFalconer said: s0****@gmail.co m wrote:
>>
... snip ...
>>
Shouldn't it be unsigned char *? Anyway, my argument was that if
your compiler can perform arithmetic with the void *, then go
void *, as you did above.
Because a compiler can't 'perform arithmetic with the void*'. It
says so, right in the C standard. Trying to do so results in
undefined behaviour.
Because it's a constraint violation, it also requires the implementation to
issue a diagnostic message. See 3.3.6 of C89 or 6.5.6 of C99.
--
Richard Heathfield <http://www.cpax.org.uk >
Email: -http://www. +rjh@
Google users: <http://www.cpax.org.uk/prg/writings/googly.php>
"Usenet is a strange place" - dmr 29 July 1999
Nick Keighley wrote:
Richard<rgr...@ gmail.comwrote:
.... snip ...
>
>My pointers were numbers. I could see them in the debugger.
but in general they aren't. Think DOS. Think IBM mainframe.
Are you trying to be dense?
No, he is just re-demonstrating the condition. I have seen many
explanations to him of this elementary fact.
--
[mail]: Chuck F (cbfalconer at maineline dot net)
[page]: <http://cbfalconer.home .att.net>
Try the download section.
On Sep 17, 5:57*pm, Keith Thompson <ks***@mib.orgw rote:
s0****@gmail.co m writes:
>On Sep 17, 2:24*pm, Eric Sosman <Er*********@su n.comwrote:
[...]
>>* * *1) The function deals with the bytes that represent an object,
* * * * not with the object as such. *I/O functions, memcpy(),
* * * * free(), and so on are type-blind in this sense: They just
* * * * care about the object's bytes, not about its nature.
>Exactly, that's the "bytes" I was referring to. The word "bytes" is
also used to refer to characters. But I was referring to the bytes
that constitute an arbitrary object: int, double, struct, etc., and
that if examined, they will appear to have meaningless values. This is
also why I find it more natural to use void *: you aren't going to
examine what it points to, because all you're going to see are some
arbitrary bits that constitute an object of (possibly) bigger size.
unsigned char is used for both kinds of "bytes".
Sure, because it's the standard way. But I think you can now see why I
find void * more natural for the above kinds of "bytes."
Sebastian s0****@gmail.co m writes:
On Sep 17, 5:57*pm, Keith Thompson <ks***@mib.orgw rote:
>s0****@gmail.co m writes:
>>On Sep 17, 2:24*pm, Eric Sosman <Er*********@su n.comwrote:
[...]
>>>* * *1) The function deals with the bytes that represent an object,
* * * * not with the object as such. *I/O functions, memcpy(),
* * * * free(), and so on are type-blind in this sense: They just
* * * * care about the object's bytes, not about its nature.
>>Exactly, that's the "bytes" I was referring to. The word "bytes" is
also used to refer to characters. But I was referring to the bytes
that constitute an arbitrary object: int, double, struct, etc., and
that if examined, they will appear to have meaningless values. This is
also why I find it more natural to use void *: you aren't going to
examine what it points to, because all you're going to see are some
arbitrary bits that constitute an object of (possibly) bigger size.
unsigned char is used for both kinds of "bytes".
Sure, because it's the standard way. But I think you can now see why I
find void * more natural for the above kinds of "bytes."
No, I really can't.
If I had the opportunity to go back in time and change C, I'd probably
separate the concepts of "byte" (the fundamental unit of storage) and
"character" . I'd probably make "byte" a keyword and a type name,
referring to a numeric type similar to what we know as unsigned char,
and I'd allow sizeof(char) to be something other than 1. Given such a
change, it would make sense to access raw memory using type byte*, and
you could sensibly perform pointer arithmetic on such a type. I'm not
certain whether void*, a type that points to raw memory but that can't
can't be dereferenced or incremented, would add sufficient value to be
worth having in the language; perhaps memcpy and friends would just
take arguments of type byte*.
But it's *way* too late to make such a change in any language calling
itself "C".
If you want to a raw pointer that can point to any arbitrary object,
use void*. If you want to access the object's representation as bytes
or perform pointer arithmetic, use unsigned char*. If you want to
point to a specific type, or to an array of a specific type, use a
pointer to that type.
You cannot legally perform pointer arithmetic on void* in standard C,
and if you attempt to do so your code will be *gratuitously*
non-portable.
--
Keith Thompson (The_Other_Keit h) ks***@mib.org <http://www.ghoti.net/~kst>
Nokia
"We must do something. This is something. Therefore, we must do this."
-- Antony Jay and Jonathan Lynn, "Yes Minister"
Richard wrote:
Chris Dollin <ch**********@h p.comwrites:
>No, I am not saying that. I'm perfectly happy to believe that your
debugger renders bits of pointer as numbers.
Thats nice. How does your debugger do it?
Probably as numbers. How would I know? It's been so long*.
When you look into memory and you see your pointer stored in an
arry how does it look? Hex number by any chance? What a surprise.
If you cast it to a char * and subtract p
from ++p do you get 4 on a 32 bit machine? I did.....
I've never looked [at this using a debugger]. I've never needed to.
I understand C's computational model for pointers, and I understand
code generation, and I understand how a /particular/ implementation
can represent pointers in the same way it represents numbers, and I
don't confuse this with a /necessary/ property of implementations ,
nor do I confuse scalars and vectors -- a segmented architectures
pointers are multi-component, like vectors, and just because you
can read off a bit-pattern and apply the traditional (but inappropriate)
binary decoding and get a number doesn't mean that bit-pattern
/is/ a number.
--
'It changed the future .. and it changed us.' /Babylon 5/
Hewlett-Packard Limited registered office: Cain Road, Bracknell,
registered no: 690597 England Berks RG12 1HN
Richard wrote:
The correct statement is
"It is incorrect to increment a non initialised variable since the
behaviour is undefined - however the value may well increment as
expected".
"Or not, at the whim of the implementor. DO NOT RELY ON THIS (if
you expect your code to be even vaguely portable)."
--
'It changed the future .. and it changed us.' /Babylon 5/
Hewlett-Packard Limited registered office: Cain Road, Bracknell,
registered no: 690597 England Berks RG12 1HN
On 17 Sep, 15:55, s0s...@gmail.co m wrote:
On Sep 17, 3:02*am,Nick Keighley<nick_k eighley_nos...@ hotmail.com>
wrote:
On 16 Sep, 21:14, s0s...@gmail.co m wrote:
>>>You have two options, do it portably, or do it with an extension, with
>>>absolutely no loss in efficiency or size, and you advice to choose the
>>>extension.
>>Well, perhaps the OP doesn't need portability (or at least not this
>>kind of portability).
there is a *well defined* and *portable* way to do this. So why choose
a non-portable way? I'm sorry I've had this sort of argument before
(in my work place) we really are on different planets. Why NOT
use a portable means, if it exists?
Because sometimes you don't need portability and because sometimes
there's a better way?
at the risk of repeating myself. I REALLY do not understand
how anyone can think like this. You wish to do something in C.
You have two options *that produce identical results* and are similar
in programming complexity. One is portable and one is not.
Why not use the portable version?
What is the <expletivepoi nt of using the non-portable version?
What can you possibly gain?
I'll probably give up after this. The trouble with text based
formats is you can't hear how hard I'm hitting my keyboard by
now.
>>Besides, void * seems more natural for this kind
>>of task.
rubbish
Why is it rubbish?
doing arithmatic on a void pointer seems really strange to me.
If you have a void pointer, it will typically point
to any kind of object: int, double, long long, structures, anything.
yup
Does it seem natural to you to treat such an object as if it were a
*character*?
no. It's not a character!
>* * *"This kind of task" was an artificial bit of code specifically
>written to experiment with manipulating pointers. *An investigative
>doodle, nothing more.
No. This "investigat e doodle" demonstrated this kind of task. But this
kind of task is also common in real-world programs, and it's
important. I, for one, have actually done this, without realizing I
was using an extension.
so now you know- so don't do it again
Why not? I'll simply keep in mind that it works only under a certain
compiler.
<nick's head explodes>
--
Nick Keighley
Consistency is the last refuge of the unimaginative.
-- Oscar Wilde
Richard<rg****@ gmail.comwrote:
Chris Dollin <ch**********@h p.comwrites:
No, I am not saying that. I'm perfectly happy to believe that your
debugger renders bits of pointer as numbers.
Thats nice. How does your debugger do it? When you look into memory and
you see your pointer stored in an arry how does it look? Hex number by
any chance? What a surprise.
My dear dunce, if you explicitly ask your debugger to hexdump
everything, _your DNA_ is a hex number. No surprise there, then.
OTOH, if I ask my debugger to do a stringdump of memory, all floating
point numbers will display as strings. Now, is a float therefore a
string?
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