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Neo
Guest
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 Posted: Mon Jan 03, 2005 7:14 am Post subject: pointer to an int array |
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Hi Folks,
#include<stdio.h>
int main()
{
int (*p)[10];
int arr[10];
int i;
p = arr; /* <-- compiler warning */
for(i=0; i <= 12; i++){ /* i <=12 delibrately written code */
*((*p) + i) = i;
printf("%dn", p[0][i]);
}
return 0;
}
In the above program compiler is giving following warning
$gcc pointer2array.c
pointer2array.c: In function `main':
pointer2array.c:8: warning: assignment from incompatible pointer type
Why the warning?
O'kay and what is the use of int (*p)[10] type of declaration ?
Does the compiler perform any check on the array bounds?
lets say if i declare int arr[12]...
-Neo
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Jack Klein
Guest
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| Posted: Mon Jan 03, 2005 7:54 am Post subject: Re: pointer to an int array |
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On Mon, 3 Jan 2005 12:44:42 +0530, "Neo" <timeless_illusion (AT) yahoo (DOT) com>
wrote in comp.lang.c:
| Quote: | Hi Folks,
#include
int main()
{
int (*p)[10];
int arr[10];
int i;
p = arr; /* <-- compiler warning */
|
The name of an array in an expression, other than when it is the
operand of the 'sizeof' or '&' operators, is converted to a pointer to
its first element. So in the statement above, you are assigning the
address of an int (arr[0]) to a pointer to an array. But arr[0] is an
int, not an array of ints.
Replace with:
p = &arr;
| Quote: | for(i=0; i <= 12; i++){ /* i <=12 delibrately written code */
*((*p) + i) = i;
printf("%dn", p[0][i]);
}
return 0;
}
In the above program compiler is giving following warning
$gcc pointer2array.c
pointer2array.c: In function `main':
pointer2array.c:8: warning: assignment from incompatible pointer type
Why the warning?
O'kay and what is the use of int (*p)[10] type of declaration ?
Does the compiler perform any check on the array bounds?
lets say if i declare int arr[12]...
|
--
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.learn.c-c++
http://www.contrib.andrew.cmu.edu/~ajo/docs/FAQ-acllc.html
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Neo
Guest
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| Posted: Mon Jan 03, 2005 8:31 am Post subject: Re: pointer to an int array |
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"Jack Klein" <jackklein (AT) spamcop (DOT) net> wrote
| Quote: | On Mon, 3 Jan 2005 12:44:42 +0530, "Neo"
wrote in comp.lang.c:
Hi Folks,
#include
int main()
{
int (*p)[10];
int arr[10];
int i;
p = arr; /* <-- compiler warning */
The name of an array in an expression, other than when it is the
operand of the 'sizeof' or '&' operators, is converted to a pointer to
its first element. So in the statement above, you are assigning the
address of an int (arr[0]) to a pointer to an array. But arr[0] is an
int, not an array of ints.
Replace with:
p = &arr;
|
O'kay, dats fine.
if I change the declaration above as :
int (*p)[]; <-- pointer2array.c:12: error: invalid use of array with
unspecified bounds
int arr[10];
p = &arr;
for(i=0; i <= 12; i++){
*((*p) + i) = i;
printf("%dn", p[0][i]);
}
compiler error!
what's the use of array index in the above declaration?
as i can go upto 12 or more...
-Neo
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Mark McIntyre
Guest
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| Posted: Mon Jan 03, 2005 11:08 am Post subject: Re: pointer to an int array |
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On Mon, 3 Jan 2005 14:01:44 +0530, in comp.lang.c , "Neo"
<timeless_illusion (AT) yahoo (DOT) com> wrote:
| Quote: | if I change the declaration above as :
int (*p)[]; <-- pointer2array.c:12: error: invalid use of array with unspecified bounds
compiler error!
|
yes, its an error
| Quote: | what's the use of array index in the above declaration?
|
none - its an illegal construct
| Quote: | as i can go upto 12 or more...
|
only by invoking undefined behaviour....
Don't do that.
--
Mark McIntyre
CLC FAQ
CLC readme: <http://www.ungerhu.com/jxh/clc.welcome.txt>
http://www.newsfeeds.com The #1 Newsgroup Service in the World! 120,000+ Newsgroups
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gooch
Guest
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| Posted: Mon Jan 03, 2005 2:10 pm Post subject: Re: pointer to an int array |
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You can index up to whatever value you want. p[1000] is fine. The
result is going to be undefined. In most cases you would get a seg
fault but that may not be the case. You have to do the bounds checking
yourself.
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Richard Bos
Guest
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| Posted: Mon Jan 03, 2005 3:36 pm Post subject: Re: pointer to an int array |
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"gooch" <gooch001 (AT) comcast (DOT) net> wrote:
[ Using Google-Broken-Beta is not an excuse not to show any context.
Learn to use it to quote properly or get a real newsreader. Context
reinstated. ]
| Quote: | "Neo" <timeless_illusion (AT) yahoo (DOT) com> wrote:
if I change the declaration above as :
int (*p)[]; <-- pointer2array.c:12: error: invalid use of array with
unspecified bounds
int arr[10];
p = &arr;
for(i=0; i <= 12; i++){
*((*p) + i) = i;
printf("%dn", p[0][i]);
}
compiler error!
what's the use of array index in the above declaration?
as i can go upto 12 or more...
You can index up to whatever value you want. p[1000] is fine. The
result is going to be undefined.
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I don't see how you can write those sentences together. The result of
p[10] and over are indeed going to be undefined, which is exactly why
such indices are _not_ fine.
| Quote: | In most cases you would get a seg fault but that may not be the case.
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If you're lucky, you'll get a segfault. If you're unlucky, you won't see
the segfault, but your customer will. If you're _really_ unlucky,
neither of you will get a segfault, but your program will slowly
scribble over all your customer's data, and he'll sue you. Don't run
over the end of your arrays, not even when you think "it'll be safe,
because you'll get a segfault anyway".
Richard
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gooch
Guest
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| Posted: Mon Jan 03, 2005 6:19 pm Post subject: Re: pointer to an int array |
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"I don't see how you can write those sentences together. ....."
I am not saying it is a good thing or that you should do it, only that
it is not a legal operation.
"If you're lucky, you'll get a segfault. ....."
Again I am not suggesting he assume a segfault will occur, only that he
needs to do the bounds checking himself.
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Mark McIntyre
Guest
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| Posted: Mon Jan 03, 2005 6:41 pm Post subject: Re: pointer to an int array |
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On 3 Jan 2005 10:19:39 -0800, in comp.lang.c , "gooch"
<gooch001 (AT) comcast (DOT) net> wrote:
(contextless stuff)
please read the VERY FIRST THING that richard said in his last post to you.
--
Mark McIntyre
CLC FAQ <http://www.eskimo.com/~scs/C-faq/top.html>
CLC readme: <http://www.ungerhu.com/jxh/clc.welcome.txt>
http://www.newsfeeds.com The #1 Newsgroup Service in the World! 120,000+ Newsgroups
----= East and West-Coast Server Farms - Total Privacy via Encryption =----
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Keith Thompson
Guest
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| Posted: Mon Jan 03, 2005 8:43 pm Post subject: Re: pointer to an int array |
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"gooch" <gooch001 (AT) comcast (DOT) net> writes:
| Quote: | You can index up to whatever value you want. p[1000] is fine. The
result is going to be undefined. In most cases you would get a seg
fault but that may not be the case. You have to do the bounds checking
yourself.
|
(I presume p is an array with fewer than 1001 elements.)
p[1000] is not "fine" in any sense of the word. It is a serious error
that the implementation is not required to diagnose. This can
sometimes have the same visible results as something that's perfectly
legal, but the distinction is critical.
Incidentally, it's not true that "in most cases you would get a seg
fault". Attempting to access an array beyond its bounds is very
likely to refer to some other variable in your program, with
unpredictable results.
Your overall point, that you have to do your own bounds checking
because the implementation won't do it for you, is quite correct; I'm
just (over)reacting to your use of the word "fine".
--
Keith Thompson (The_Other_Keith) [email]kst-u (AT) mib (DOT) org[/email] <http://www.ghoti.net/~kst>
San Diego Supercomputer Center <*> <http://users.sdsc.edu/~kst>
We must do something. This is something. Therefore, we must do this.
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Stan Milam
Guest
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| Posted: Tue Jan 04, 2005 5:15 am Post subject: Re: pointer to an int array |
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Jack Klein wrote:
| Quote: | On Mon, 3 Jan 2005 12:44:42 +0530, "Neo"
wrote in comp.lang.c:
Hi Folks,
#include
int main()
{
int (*p)[10];
int arr[10];
int i;
p = arr; /* <-- compiler warning */
The name of an array in an expression, other than when it is the
operand of the 'sizeof' or '&' operators, is converted to a pointer to
its first element. So in the statement above, you are assigning the
address of an int (arr[0]) to a pointer to an array. But arr[0] is an
int, not an array of ints.
Replace with:
p = &arr;
for(i=0; i <= 12; i++){ /* i <=12 delibrately written code */
*((*p) + i) = i;
printf("%dn", p[0][i]);
}
return 0;
}
In the above program compiler is giving following warning
$gcc pointer2array.c
pointer2array.c: In function `main':
pointer2array.c:8: warning: assignment from incompatible pointer type
Why the warning?
O'kay and what is the use of int (*p)[10] type of declaration ?
Does the compiler perform any check on the array bounds?
lets say if i declare int arr[12]...
|
Okay, both of you are mistaken. int (*p)[10] is declaring an array of
10 pointers to function that return integers. Not quite the same thing
as either of you are talking about.
--
Regards,
Stan Milam.
-----------------------------------------------------------------------------
Vita Brevis. Carpe Guitarum! - Jamie Kinscherff
-----------------------------------------------------------------------------
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Neo
Guest
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| Posted: Tue Jan 04, 2005 6:21 am Post subject: Re: pointer to an int array |
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"Stan Milam" <stmilam (AT) swbell (DOT) net> wrote
| Quote: | Jack Klein wrote:
On Mon, 3 Jan 2005 12:44:42 +0530, "Neo"
wrote in comp.lang.c:
Hi Folks,
#include
int main()
{
int (*p)[10];
int arr[10];
int i;
p = arr; /* <-- compiler warning */
The name of an array in an expression, other than when it is the
operand of the 'sizeof' or '&' operators, is converted to a pointer to
its first element. So in the statement above, you are assigning the
address of an int (arr[0]) to a pointer to an array. But arr[0] is an
int, not an array of ints.
Replace with:
p = &arr;
for(i=0; i <= 12; i++){ /* i <=12 delibrately written code */
*((*p) + i) = i;
printf("%dn", p[0][i]);
}
return 0;
}
In the above program compiler is giving following warning
$gcc pointer2array.c
pointer2array.c: In function `main':
pointer2array.c:8: warning: assignment from incompatible pointer type
Why the warning?
O'kay and what is the use of int (*p)[10] type of declaration ?
Does the compiler perform any check on the array bounds?
lets say if i declare int arr[12]...
Okay, both of you are mistaken. int (*p)[10] is declaring an array of 10
pointers to function that return integers. Not quite the same thing as
either of you are talking about.
--
Regards,
Stan Milam.
-----------------------------------------------------------------------------
Vita Brevis. Carpe Guitarum! - Jamie Kinscherff
-----------------------------------------------------------------------------
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pointers to function???
hmm.... its a pointer to arrary of 10 INTs!
-Neo
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Jens.Toerring@physik.fu-b
Guest
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| Posted: Tue Jan 04, 2005 8:19 am Post subject: Re: pointer to an int array |
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Stan Milam <stmilam (AT) swbell (DOT) net> wrote:
| Quote: | Okay, both of you are mistaken. int (*p)[10] is declaring an array of
10 pointers to function that return integers. Not quite the same thing
as either of you are talking about.
|
Sorry, but no.
int ( *p )[ 10 ];
is a pointer to an array of 10 ints. The parentheses here are needed to
distinguish it from
int *p[ 10 ];
which would be an array of 10 pointers to int. An array of 10 pointers
to functions that return int would be
int ( *p[ 10 ] )( );
Regards, Jens
--
Jens Thoms Toerring ___ [email]Jens.Toerring (AT) physik (DOT) fu-berlin.de[/email]
__________________________ http://www.toerring.de
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Lawrence Kirby
Guest
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| Posted: Tue Jan 04, 2005 2:07 pm Post subject: Re: pointer to an int array |
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On Mon, 03 Jan 2005 12:44:42 +0530, Neo wrote:
| Quote: | Hi Folks,
#include
int main()
{
int (*p)[10];
int arr[10];
int i;
p = arr; /* <-- compiler warning */
for(i=0; i <= 12; i++){ /* i <=12 delibrately written code */
*((*p) + i) = i;
printf("%dn", p[0][i]);
}
return 0;
}
In the above program compiler is giving following warning
$gcc pointer2array.c
pointer2array.c: In function `main':
pointer2array.c:8: warning: assignment from incompatible pointer type
Why the warning?
|
Becuase you are trying to assign a value of type int * to a variable of
type int (*)[10]. C doesn't support implicit conversion between these
types. You could write
p = &arr;
but there isn't anything obvious to be gained over defining p as int *.
| Quote: | O'kay and what is the use of int (*p)[10] type of declaration ?
|
You could use this in something like the following
int arr2[5][10];
int (*p2)[10] = arr2;
Then accessing p2[x][y] will access the element arr2[x][y]
| Quote: | Does
the compiler perform any check on the array bounds? lets say if i
declare int arr[12]...
|
C doesn't require any sort of bounds checking from the compiler, and most
don't attempt to.
Lawrence
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S.Tobias
Guest
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| Posted: Fri Jan 07, 2005 10:31 pm Post subject: Re: pointer to an int array |
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Mark McIntyre <markmcintyre (AT) spamcop (DOT) net> wrote:
| Quote: | On Mon, 3 Jan 2005 14:01:44 +0530, in comp.lang.c , "Neo"
[email]timeless_illusion (AT) yahoo (DOT) com[/email]> wrote:
if I change the declaration above as :
int (*p)[]; <-- pointer2array.c:12: error: invalid use of array with unspecified bounds
compiler error!
yes, its an error
|
IMHO I think you're wrong here. `p' is simply a pointer to incomplete type
("array of int of unknown size").
The OP probably hadn't noticed that the error was at different line:
printf("%dn", p[0][i]); //error
Exactly, the offending expression is p[0], because, it gets translated
to:
*(p + 0)
in which the summation cannot be evaluated because `*p' has an incomplete
type (however I fail to explain exactly why, because this expression doesn't
seem to violate any constraints at the "+" operator; could anyone help?).
The above expression can be corrected with:
(*p)[i];
This is interesting, because we seem to apply indirection operator to
pointer to incomplete type (again, this is not a constraint violation,
however it doesn't work with pointers to incomplete struct type; could
someone help explain this please?); but here lvalue `*p', which is
of "array of int" type, is converted to "pointer to int" and points
to its first element, essentially same location as the array itself;
this rule seems to take precedence.
| Quote: | what's the use of array index in the above declaration?
|
Type `int[]' is compatible with `int[10]' and `int[20]',
so `p' could be assigned an address of objects defined as:
int a[15];
int b[80];
double d[10]; //except this object
p = &a;
p = &b;
p = &d; //error
However, since `p' is a pointer to incomplete type, no arithmetic or
comparisons can be made with it.
I think one can describe `p' as a generic pointer to "arrays of int".
I don't know if it could be more useful than that.
| Quote: | as i can go upto 12 or more...
only by invoking undefined behaviour....
|
Yes.
--
Stan Tobias
mailx `echo [email]siXtY (AT) FamOuS (DOT) BedBuG.pAlS.INVA[/email]LID | sed s/[[:upper:]]//g`
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Lawrence Kirby
Guest
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| Posted: Tue Jan 11, 2005 4:01 pm Post subject: Re: pointer to an int array |
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On Fri, 07 Jan 2005 22:31:16 +0000, S.Tobias wrote:
| Quote: | Mark McIntyre <markmcintyre (AT) spamcop (DOT) net> wrote:
On Mon, 3 Jan 2005 14:01:44 +0530, in comp.lang.c , "Neo"
[email]timeless_illusion (AT) yahoo (DOT) com[/email]> wrote:
if I change the declaration above as :
int (*p)[]; <-- pointer2array.c:12: error: invalid use of array with unspecified bounds
compiler error!
yes, its an error
IMHO I think you're wrong here. `p' is simply a pointer to incomplete type
("array of int of unknown size").
The OP probably hadn't noticed that the error was at different line:
printf("%dn", p[0][i]); //error
Exactly, the offending expression is p[0], because, it gets translated
to:
*(p + 0)
in which the summation cannot be evaluated because `*p' has an incomplete
type (however I fail to explain exactly why, because this expression doesn't
seem to violate any constraints at the "+" operator; could anyone help?).
|
In both C90 and C99 the + operator requires a pointer operand to have a
pointer to an object type; an incomplete type is not an object type. This
is a constraint.
Lawrence
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