What is wrong with the above?
Don't worry, I already know (learned my lesson last week.) It is for the
benefit of our resident compiler guru who seems to think you need the cast.
I thought it too, up until I started posting here!
Thanks,
Chris
Nov 14 '05
33  2726 
Chris Fogelklou <ch************ *@comhem.se> wrote: int my_func(void *pinst)
{
my_struct *pthis = (my_struct *)pinst; //Cast
pthis->print("Lets see what's wrong with this!");
return 0;
}
If it is OK to never cast to/from void * as was made abundantly clear last
week, then this is also OK. However, pinst could be a run-time defineable
value.
It is absolutly ok to do this (presuming a correct definition of
my_struct) with or without a cast.
void * can be converted to from other object pointer types and the other
way round without casts in any circumstance. (function pointers are
different here, but that is another story).
And note that function parameters are always evaluated at runtime.
When my_func is called, the arguments to that call are evaluated and
copied into the parameters of the to be called function.
--
Z (Zo**********@d aimlerchrysler. com)
"LISP is worth learning for the profound enlightenment experience
you will have when you finally get it; that experience will make you
a better programmer for the rest of your days." -- Eric S. Raymond
"Chris Fogelklou" <ch************ *@comhem.se> wrote in message
news:o3******** ************@ne wsc.telia.net.. . "Thomas Stegen" <ts*****@cis.st rath.ac.uk> wrote in message
news:c5******** ****@ID-228872.news.uni-berlin.de... Chris Fogelklou wrote:
What is wrong with the above?
Don't worry, I already know (learned my lesson last week.) It is for the benefit of our resident compiler guru who seems to think you need the cast. I thought it too, up until I started posting here!
struct my_struct *p = (struct my_struct *)malloc(sizeof (struct
my_struct));
There is nothing wrong with the above. It is considered a better
idiom to use
struct my_struct *p = malloc(sizeof *p);
The argument that the compiler does not know the type of the object
is bogus since all the information is already there. The compiler knows
the type of p, it knows the type of the value returned by malloc,
nothing is missing.
In my previous reply I mentioned 6.3.2.3, I should also have mentioned
clause 6.5.4 and 6.5.16.1 which is where the "no cast required" part is.
--
Thomas.
Hi Thomas,
Thanks! OK, then what about this:
int my_func(void *pinst)
{
my_struct *pthis = (my_struct *)pinst; //Cast
pthis->print("Lets see what's wrong with this!");
return 0;
}
If it is OK to never cast to/from void * as was made abundantly clear last
week, then this is also OK. However, pinst could be a run-time defineable
value.
Thanks!
Chris
Sorry... meant "then this is also BAD practice."
"CBFalconer " <cb********@yah oo.com> wrote in message
news:40******** *******@yahoo.c om... Chris Fogelklou wrote: "Chris Fogelklou" <ch************ *@comhem.se> wrote in message
What is wrong with the above?
Don't worry, I already know (learned my lesson last week.) It
is for the benefit of our resident compiler guru who seems to
think you need the cast. I thought it too, up until I started
posting here!
Even better, here is the email itself. Please help!
<start email>
First, Casts from void * are necessary both in C and C++. This is
from the same reason you told me: otherwise, if the compiler
wouldn't know what is the size of the object which is pointed at,
it might get confused with arithmetic operations. You can ask him
where did he learned this rule from. tell me if you want me to
correspond directly with him.
Second, a possible reason to use integral types for addresses
rather than void *, is that in order to perform bitwise operations
on an address, you have to use operators of an integral type.
<end email>
Don't get too antsy with him/her. This sounds like some system
programming operations on some embedded system, where the rules
are often different. The cast of the malloc remains unnecessary
(and don't put the question in the subject only).
Do learn to organize your questions into a single consistent
message, rather than a bit here and a bit there.
Don't worry... I'm not getting antsy (lots of smileys in my email to her).
But she's the expert and I'm just the user. Good guess that it is an
embedded system; that it is!
It's easier to ask here and get exact references than to say "but I
think..." (I made the "but I think" mistake on this NG last week :)
Next time I'll put the question in the post as well.
Cheers, Chris
In <c5************ @ID-228872.news.uni-berlin.de> Thomas Stegen <ts*****@cis.st rath.ac.uk> writes: Chris Fogelklou wrote:
Even better, here is the email itself. Please help!
<start email>
First, Casts from void * are necessary both in C and C++. This is from the
same reason you told me: otherwise, if the compiler wouldn't
know what is the size of the object which is pointed at, it might get
confused with arithmetic operations. You can ask him where did he learned
this rule from. tell me if you want me to correspond directly with him.
Yes, please tell her to come here. On the other hand, don't, tell
her to read the archives. This conversation is getting old ;)
the ISO/IEC 9899:1999 standard section 6.3.2.3 has what you want.
Nope, it doesn't. It says:
1 A pointer to void may be converted to or from a pointer to any
incomplete or object type. A pointer to any incomplete or
object type may be converted to a pointer to void and back again;
the result shall compare equal to the original pointer.
But there is no mention that the conversion is automatic. Compare with
5 An integer may be converted to any pointer type. Except as
previously specified, the result is implementation-defined,
might not be correctly aligned, might not point to an entity of
the referenced type, and might be a trap representation.
6 Any pointer type may be converted to an integer type. Except as
previously specified, the result is implementation-defined. If the
result cannot be represented in the integer type, the behavior
is undefined. The result need not be in the range of values of
any integer type.
7 A pointer to an object or incomplete type may be converted to
a pointer to a different object or incomplete type.
Do you mean that these conversions are automatic, too?
Dan
--
Dan Pop
DESY Zeuthen, RZ group
Email: Da*****@ifh.de
In <40************ ***@yahoo.com> CBFalconer <cb********@yah oo.com> writes: Chris Fogelklou wrote: "Chris Fogelklou" <ch************ *@comhem.se> wrote in message
> What is wrong with the above?
>
> Don't worry, I already know (learned my lesson last week.) It
> is for the benefit of our resident compiler guru who seems to
> think you need the cast. I thought it too, up until I started
> posting here!
>
Even better, here is the email itself. Please help!
<start email>
First, Casts from void * are necessary both in C and C++. This is
from the same reason you told me: otherwise, if the compiler
wouldn't know what is the size of the object which is pointed at,
it might get confused with arithmetic operations. You can ask him
where did he learned this rule from. tell me if you want me to
correspond directly with him.
Second, a possible reason to use integral types for addresses
rather than void *, is that in order to perform bitwise operations
on an address, you have to use operators of an integral type.
<end email>
Don't get too antsy with him/her. This sounds like some system
programming operations on some embedded system, where the rules
are often different.
It sounds more like someone confusing the C and C++ rules.
Dan
--
Dan Pop
DESY Zeuthen, RZ group
Email: Da*****@ifh.de
In <c5**********@n ews-reader4.wanadoo .fr> "Régis Troadec" <re**@wanadoo.f r> writes: "Chris Fogelklou" <ch************ *@comhem.se> a écrit dans le message de
news:3H******* *************@n ewsc.telia.net. ..
You're WRONG!
void* is implicitly converted to the destination pointer type (C99 §6.3.2.3
and §7.20.3.1).
You're WRONG!
There is NOTHING in C99 §6.3.2.3 or §7.20.3.1 supporting your assertion.
Dan
--
Dan Pop
DESY Zeuthen, RZ group
Email: Da*****@ifh.de
"Dan Pop" <Da*****@cern.c h> wrote in message
news:c5******** **@sunnews.cern .ch... In <c5************ @ID-228872.news.uni-berlin.de> Thomas Stegen
<ts*****@cis.st rath.ac.uk> writes:Chris Fogelklou wrote:
Even better, here is the email itself. Please help!
<start email>
First, Casts from void * are necessary both in C and C++. This is from
the same reason you told me: otherwise, if the compiler wouldn't
know what is the size of the object which is pointed at, it might get
confused with arithmetic operations. You can ask him where did he
learned this rule from. tell me if you want me to correspond directly with him.
Yes, please tell her to come here. On the other hand, don't, tell
her to read the archives. This conversation is getting old ;)
the ISO/IEC 9899:1999 standard section 6.3.2.3 has what you want.
Nope, it doesn't. It says:
1 A pointer to void may be converted to or from a pointer to any
incomplete or object type. A pointer to any incomplete or
object type may be converted to a pointer to void and back again;
the result shall compare equal to the original pointer.
But there is no mention that the conversion is automatic. Compare with
5 An integer may be converted to any pointer type. Except as
previously specified, the result is implementation-defined,
might not be correctly aligned, might not point to an entity of
the referenced type, and might be a trap representation.
6 Any pointer type may be converted to an integer type. Except as
previously specified, the result is implementation-defined. If the
result cannot be represented in the integer type, the behavior
is undefined. The result need not be in the range of values of
any integer type.
7 A pointer to an object or incomplete type may be converted to
a pointer to a different object or incomplete type.
Do you mean that these conversions are automatic, too?
Looks to me like if 1.) is, then 7.) is as well. Both 5.) and 6.) claim
that the result is implementation defined and the behaviour might be
undefined.
Dan Pop wrote: In <c5************ @ID-228872.news.uni-berlin.de> Thomas Stegen <ts*****@cis.st rath.ac.uk> writes:
Nope, it doesn't. It says:
I gave the other relevant sections in a different post.
--
Thomas.
Chris Fogelklou wrote:
Looks to me like if 1.) is, then 7.) is as well. Both 5.) and 6.) claim
that the result is implementation defined and the behaviour might be
undefined.
The right section to look at would be 6.5.4 and 6.5.16.1.
--
Thomas.
"Dan Pop" <Da*****@cern.c h> a écrit dans le message de
news:c5******** **@sunnews.cern .ch... In <c5**********@n ews-reader4.wanadoo .fr> "Régis Troadec"
<re**@wanadoo.f r> writes:
"Chris Fogelklou" <ch************ *@comhem.se> a écrit dans le message de
news:3H******* *************@n ewsc.telia.net. ..
You're WRONG!
void* is implicitly converted to the destination pointer type (C99
§6.3.2.3and §7.20.3.1).
You're WRONG!
There is NOTHING in C99 §6.3.2.3 or §7.20.3.1 supporting your assertion.
Let's correct it : void * may be converted to a pointer to any incomplete or
object type. As well, void* may be converted from a pointer to any
incomplete or object type. This is what I read in C99 §6.3.2.3.
You were probably hurt by seeing "void* IS ... converted".
Regis
Dan
--
Dan Pop
DESY Zeuthen, RZ group
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