Velocity Reviews > Initializing Pointer to an array

# Initializing Pointer to an array

rupesh_533@rediffmail.com
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 10-20-2005
I am assuming the following things.
1.Pointer to an integer means it points to an integer,On incrementing
the pointer,it points to the next integer in memory.
2.Pointer to an array of some size means it points to an array,On
incrementing the pointer,it should point to next array of that size.

Correct me if i am wrong.
We declare a pointer to an integer array of Size N as
int (*a)[N];

Now My Problem is
I have an array of 100 Integers,Pointer to an array of 10 integers
int (*a)[10];
int b[100];
How can i initialize the pointer a to point to the array b base

rupesh_533@rediffmail.com
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Posts: n/a

 10-20-2005
I am able to initialize the pointer to the array as
int (*a)[10];
int b[100];

a = &b[0];
But i am getting the warning
warning: assignment from incompatible pointer type

My Code is
int main()
{
int (*a)[10];
int b[100];
int i;

for(i=0;i<100;i++)
b[i]=i;

a = &b[0];

printf(" %u %u %d %d\n",a ,&b[0],**a,b[0]);
a++;
printf(" %u %u %d %d\n",a ,&b[10],**a,b[10]);
a++;
printf(" %u %u %d %d\n",a , &b[20],**a,b[20]);
a++;
printf(" %u %u %d %d\n",a , &b[30],**a,b[30]);
}

I am getting correct results as
3221219184 3221219184 0 0
3221219224 3221219224 10 10
3221219264 3221219264 20 20
3221219304 3221219304 30 30

Christopher Benson-Manica
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Posts: n/a

 10-20-2005
http://www.velocityreviews.com/forums/(E-Mail Removed) wrote:

> I am able to initialize the pointer to the array as
> int (*a)[10];
> int b[100];

> a = &b[0];
> But i am getting the warning
> warning: assignment from incompatible pointer type

Of course you are. a is a pointer to an array of 10 ints (which
you probably do not need), while &b[0] is a pointer to an int. Your
compiler is doing you a favor.

> My Code is
> int main()

int main( void ) /* better */

> {
> int (*a)[10];
> int b[100];
> int i;

> for(i=0;i<100;i++)

for( i=0; i < sizeof b; i++) /* better */

> b[i]=i;

> a = &b[0];

Wrong, as I said.

> printf(" %u %u %d %d\n",a ,&b[0],**a,b[0]);

Wrong in multiple ways:

1) You did not include <stdio.h>.
2) %u is not the conversion specifier you want for pointers.

printf( " %p %p %d %d\n", (void*)a, (void*)&b[0], **a, b[0] );

Note the casts; they are required.

> a++;
> printf(" %u %u %d %d\n",a ,&b[10],**a,b[10]);
> a++;
> printf(" %u %u %d %d\n",a , &b[20],**a,b[20]);
> a++;
> printf(" %u %u %d %d\n",a , &b[30],**a,b[30]);

Unless you are using a C99 compiler, you must return a value from
omission.

> }

--
Christopher Benson-Manica | I *should* know what I'm talking about - if I
ataru(at)cyberspace.org | don't, I need to know. Flames welcome.

Antonio Contreras
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Posts: n/a

 10-20-2005
Christopher Benson-Manica wrote:

<snipped rest of code>

> (E-Mail Removed) wrote:
> > int main()

>
> int main( void ) /* better */
>
> > {
> > int (*a)[10];
> > int b[100];
> > int i;

>
> > for(i=0;i<100;i++)

>
> for( i=0; i < sizeof b; i++) /* better */
>

Shouldn't that be:
for (i = 0; i < ((sizeof b) / (sizeof int)); i++)
??

Christopher Benson-Manica
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 10-20-2005
Antonio Contreras <(E-Mail Removed)> wrote:

> for (i = 0; i < ((sizeof b) / (sizeof int)); i++)

Yes. Ouch. (And thank you.)

--
Christopher Benson-Manica | I *should* know what I'm talking about - if I
ataru(at)cyberspace.org | don't, I need to know. Flames welcome.

Flash Gordon
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 10-20-2005
Christopher Benson-Manica wrote:
> Antonio Contreras <(E-Mail Removed)> wrote:
>
>>for (i = 0; i < ((sizeof b) / (sizeof int)); i++)

>
> Yes. Ouch. (And thank you.)

Or better yet:
for (i = 0; i < ((sizeof b) / (sizeof *b)); i++)

Then it will compile (apart from being incomplete and the certainty of
me introducing an error in my correction) and be independant of the type
of array b.

--
Flash Gordon
Living in interesting times.
Although my email address says spam, it is real and I read it.

Antonio Contreras
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Posts: n/a

 10-20-2005
Flash Gordon wrote:
> Christopher Benson-Manica wrote:
> > Antonio Contreras <(E-Mail Removed)> wrote:
> >
> >>for (i = 0; i < ((sizeof b) / (sizeof int)); i++)

> >
> > Yes. Ouch. (And thank you.)

>
> Or better yet:
> for (i = 0; i < ((sizeof b) / (sizeof *b)); i++)
>
> Then it will compile (apart from being incomplete and the certainty of
> me introducing an error in my correction) and be independant of the type
> of array b.
>
>

Well, for purely aesthetic reasons I would prefer the totally
equivalent:

for (i = 0; i < ((sizeof b) / (sizeof b[0])); i++)

Call me paranoid, but the equivalence between arrays and pointers is
confusing enough [1] without dereferencing pointers that have been
created by the decay of an array name.

[1] Actually I've gotten used to it and it's not confusing anymore, but
it was really confusing two years ago.

Michael Wojcik
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 10-20-2005

In article <(E-Mail Removed) .com>, "Antonio Contreras" <(E-Mail Removed)> writes:
> Christopher Benson-Manica wrote:
> > (E-Mail Removed) wrote:
> > > int (*a)[10];
> > > int b[100];
> > > int i;

> >
> > > for(i=0;i<100;i++)

> >
> > for( i=0; i < sizeof b; i++) /* better */

>
> Shouldn't that be:
> for (i = 0; i < ((sizeof b) / (sizeof int)); i++)

When the sizeof operator is applied to a type, the name of the type
must be parenthesized. You're missing parentheses around "int"
(which is ironic, considering all the unnecessary sets of parentheses
you have there).

What would be better is

for (i = 0; i < sizeof b / sizeof *b; i++)

which works for a "b" of any (complete) array type.

--
Michael Wojcik (E-Mail Removed)

Global warming is just a theory. This is Intelligent Defrosting. -- "Gregg"

Peter Nilsson
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 10-20-2005
Christopher Benson-Manica wrote:
>
> Unless you are using a C99 compiler, you must return a value from
> main().

It's not mandatory un C90, just desirable.

> omission.

Perhaps, but unlike C++, both C90 and C99 allow non-void functions to
fail
to return a value. So long as the calling function doesn't attempt to
use
the value, all is fine.

As this is a <cough> feature of the language, many old compilers won't
issue a warning for this.

--
Peter

Jack Klein
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Posts: n/a

 10-21-2005
On 20 Oct 2005 15:36:14 -0700, "Antonio Contreras" <(E-Mail Removed)>
wrote in comp.lang.c:

> Flash Gordon wrote:
> > Christopher Benson-Manica wrote:
> > > Antonio Contreras <(E-Mail Removed)> wrote:
> > >
> > >>for (i = 0; i < ((sizeof b) / (sizeof int)); i++)
> > >
> > > Yes. Ouch. (And thank you.)

> >
> > Or better yet:
> > for (i = 0; i < ((sizeof b) / (sizeof *b)); i++)
> >
> > Then it will compile (apart from being incomplete and the certainty of
> > me introducing an error in my correction) and be independant of the type
> > of array b.
> >
> >

>
> Well, for purely aesthetic reasons I would prefer the totally
> equivalent:
>
> for (i = 0; i < ((sizeof b) / (sizeof b[0])); i++)
>
> Call me paranoid, but the equivalence between arrays and pointers is
> confusing enough [1] without dereferencing pointers that have been
> created by the decay of an array name.

The sizeof operator never evaluates the value of its operand in C
prior to C99, and only to determine the size of a variable length
array in C99. Under no circumstances does sizeof dereference a
pointer given to it as an argument.

Plus the C language absolutely requires that the expression b[0] be
evaluated as if written *b, so the Christopher's code and your
alternative are exactly equivalent in C.

> [1] Actually I've gotten used to it and it's not confusing anymore, but
> it was really confusing two years ago.

In either case, there is no evaluation of the value of b[0] or *b, and
no dereference.

--
Jack Klein
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