hello, this is a piece of code ,which is giving an error.
#include<stdio. h>
int main()
{
int a =10;
void *p = &a;
printf("%d ", *p ); //error....why should it //be an error ?can't the
compiler make out because //of %d that the pointer is supposed to refer to
an integer ?? or is explicit type casting required ??
++p ; //agree that this is an error.
return 0;
}
plssss tell me about the first error.thanx.
ranjan.
56  3345 
maadhuu wrote: hello, this is a piece of code ,which is giving an error.
#include<stdio. h>
int main()
{
int a =10;
void *p = &a;
printf("%d ", *p ); //error....why should it //be an error ?can't the
compiler make out because //of %d that the pointer is supposed to refer to
an integer ?? or is explicit type casting required ??
*p is illegal because you are not permitted to dereference a void
pointer. That you clearly "intended" *p to be an int doesn't permit
the compiler to ignore the error.
If *someVoidPointe r *were* legal, the result would be the unique
value of the void type [1], which isn't an int either.
[1] No, I don't subscribe to the nonsense that `void` is an empty type.
--
Chris "electric hedgehog" Dollin
predicting self-predictors' predictions is predictably unpredictable.
Chris Dollin wrote: maadhuu wrote:
hello, this is a piece of code ,which is giving an error.
#include<stdio. h>
int main()
{
int a =10;
void *p = &a;
printf("%d ", *p ); //error....why should it //be an error ?can't the
compiler make out because //of %d that the pointer is supposed to refer to
an integer ?? or is explicit type casting required ??
*p is illegal because you are not permitted to dereference a void
pointer. That you clearly "intended" *p to be an int doesn't permit
the compiler to ignore the error.
In this case the printf statement is incorrect, but simply
dereferencing a valid void pointer is not illegal.
If *someVoidPointe r *were* legal
Which it is...
the result would be the unique
value of the void type [1], which isn't an int either.
It yields an expression of type void. Not sure what you mean by unique
value of void type.
[1] No, I don't subscribe to the nonsense that `void` is an empty type.
Oh, not one of those suckers who blindly subscribes to the Standard?
6.2.5p19:
"The void type comprises an empty set of values; it is an incomplete
type that cannot be completed."
Robert Gamble
maadhuu <ma************ @yahoo.com> wrote: int main()
{
int a =10;
void *p = &a;
printf("%d ", *p );
can't the compiler make out because of %d that the pointer is supposed
to refer to an integer?
No.
++p; //agree that this is an error.
The same principle applies; if the compiler knew what p pointed at,
then both this and *p would be acceptable.
or is explicit type casting required ??
Yes.
--
Christopher Benson-Manica | I *should* know what I'm talking about - if I
ataru(at)cybers pace.org | don't, I need to know. Flames welcome.
Robert Gamble wrote: Chris Dollin wrote: maadhuu wrote:
> hello, this is a piece of code ,which is giving an error.
> #include<stdio. h>
>
> int main()
> {
> int a =10;
> void *p = &a;
> printf("%d ", *p ); //error....why should it //be an error ?can't
> the
> compiler make out because //of %d that the pointer is supposed to refer
> to an integer ?? or is explicit type casting required ??
*p is illegal because you are not permitted to dereference a void
pointer. That you clearly "intended" *p to be an int doesn't permit
the compiler to ignore the error.
In this case the printf statement is incorrect, but simply
dereferencing a valid void pointer is not illegal.
If *someVoidPointe r *were* legal
Which it is...
I beg to differ. It is not legal in standard C. Your very own quote
below says that `void` is an incomplete type, and incomplete types
cannot be dereferenced when I last looked.
the result would be the unique
value of the void type [1], which isn't an int either.
It yields an expression of type void. Not sure what you mean by unique
value of void type.
Evaluating an expression of type T returns a result which is an
instance of T -- eg, `1+2` is an expression of type `int` who's
result is the value `3`, an instance of `int`.
The unique value of type `void` is the single, unique value which
is of type `void`.
[1] No, I don't subscribe to the nonsense that `void` is an empty type.
Oh, not one of those suckers who blindly subscribes to the Standard?
That's right.
6.2.5p19:
"The void type comprises an empty set of values; it is an incomplete
type that cannot be completed."
Yes, I know. That's why I said that I didn't subscribe to that
nonsense; because I don't. The view that `void` is an empty type
leads to unfortunate conclusions about the existance of functions
returning void, which are trivially avoided by making void have
a unique instance.
--
Chris "electric hedgehog" Dollin
predicting self-predictors' predictions is predictably unpredictable.
Chris Dollin <ke**@hpl.hp.co m> wrote: Robert Gamble wrote:
Chris Dollin wrote: [1] No, I don't subscribe to the nonsense that `void` is an empty type.
6.2.5p19:
"The void type comprises an empty set of values; it is an incomplete
type that cannot be completed."
Yes, I know. That's why I said that I didn't subscribe to that
nonsense; because I don't. The view that `void` is an empty type
leads to unfortunate conclusions about the existance of functions
returning void, which are trivially avoided by making void have
a unique instance.
I think you have that backwards. If void had a unique instance, void
functions would return that instance, which could be used or assigned to
something. Since void is empty and does _not_ have any values, a void
function returns nothing, not even a unique instance, and you can't
assign nothing to an object.
Richard
Chris Dollin wrote: Robert Gamble wrote:
Chris Dollin wrote: maadhuu wrote:
> hello, this is a piece of code ,which is giving an error.
> #include<stdio. h>
>
> int main()
> {
> int a =10;
> void *p = &a;
> printf("%d ", *p ); //error....why should it //be an error ?can't
> the
> compiler make out because //of %d that the pointer is supposed to refer
> to an integer ?? or is explicit type casting required ??
*p is illegal because you are not permitted to dereference a void
pointer. That you clearly "intended" *p to be an int doesn't permit
the compiler to ignore the error.
In this case the printf statement is incorrect, but simply
dereferencing a valid void pointer is not illegal.
If *someVoidPointe r *were* legal
Which it is...
I beg to differ. It is not legal in standard C. Your very own quote
below says that `void` is an incomplete type, and incomplete types
cannot be dereferenced when I last looked.
void is an incomplete type, a pointer to void is not. We are talking
about dereferencing a pointer to void.
the result would be the unique
value of the void type [1], which isn't an int either.
It yields an expression of type void. Not sure what you mean by unique
value of void type.
Evaluating an expression of type T returns a result which is an
instance of T -- eg, `1+2` is an expression of type `int` who's
result is the value `3`, an instance of `int`.
The unique value of type `void` is the single, unique value which
is of type `void`.
That makes no sense.
[1] No, I don't subscribe to the nonsense that `void` is an empty type.
Oh, not one of those suckers who blindly subscribes to the Standard?
That's right.
Well there is little hope for you then. You talk about what is "legal
in standard C" and then admit that you don't accept the Standard itself
making your entire argument completely pointless.
6.2.5p19:
"The void type comprises an empty set of values; it is an incomplete
type that cannot be completed."
Yes, I know. That's why I said that I didn't subscribe to that
nonsense; because I don't.
If you think the Standard is nonsense then you have no ground to argue
what is or is not Standard complient.
The view that `void` is an empty type
leads to unfortunate conclusions about the existance of functions
returning void, which are trivially avoided by making void have
a unique instance.
You are going to have to elaborate on this.
Robert Gamble
printf is not part of the C language, its part of the library. The
compiler has no idea what %d in your string is supposed to mean. It
simply gets passed and interpretted at runtime. The C compiler doesn't
"look into" strings, nor what they are getting passed to.
Remember, printf could be overridden by your own code. You could
redefine printf to do something completely different and link it into
your code, which does not do anything at all with %d. printf is simply
a library function, not a part of the language itself.
Jon
----
Learn to program using Linux assembly language http://www.cafeshops.com/bartlettpublish.8640017
Jonathan Bartlett <jo*****@eskimo .com> wrote:
[ Do not post without content. Google are bad enough netizens to do so,
but this does not mean you should emulate them. ] printf is not part of the C language, its part of the library.
That distinction is meaningless. printf() is part of a correct C
implementation.
The compiler has no idea what %d in your string is supposed to mean.
The compiler is _allowed_ to have no idea by the Standard. Passing
arguments to printf() that don't correspond with the format string
invokes undefined behaviour, which means that the implementation may or
may not notice, as it pleases. Had the Standard made this a constraint
violation, the implementation would have been required to diagnose this,
no matter which distinction between "compiler" and "library" many of
them currently make.
The C compiler doesn't "look into" strings,
This is misleading. For example, it certainly does look into them to
check for escape sequences. It is certainly also allowed to look into
printf()'s format string, and check the rest of its arguments; in fact,
some implementations do so.
Remember, printf could be overridden by your own code.
No, it couldn't. Again, _some_ implementations allow this; but in ISO C,
trying to override the Standard functions invokes undefined behaviour.
And in fact, it can have disastrous results if the rest of the library
happens to rely on that one function being the one they expect.
You could redefine printf to do something completely different and link
it into your code, which does not do anything at all with %d.
Not if you want it to run anywhere except on the implementation you
wrote it on, you can't.
printf is simply a library function, not a part of the language itself.
Again, in ISO C this distinction is to all intents and purposes devoid
of meaning. Either you have an ISO C implementation, or you do not.
Richard
Richard Bos <rl*@hoekstra-uitgeverij.nl> wrote: [ Do not post without content. Google are bad enough netizens to do so,
but this does not mean you should emulate them. ]
ITYM "context"; only Skybuck posts without content :-)
--
Christopher Benson-Manica | I *should* know what I'm talking about - if I
ataru(at)cybers pace.org | don't, I need to know. Flames welcome. This thread has been closed and replies have been disabled. Please start a new discussion. Similar topics | |
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