pointer and array help

[ali]

Hi,

I'm trying to understand the reason for different output on the
following codes

Code1:

#include <iostream.h>

int main()
{
int array[]={2,4,5};

int *pointer =0;

pointer=array;

cout<<pointer<<endl;

return 0;
}
Code2:

#include <iostream.h>

int main()
{
char array[]="words";

char *pointer =0;

pointer=array;

cout<<pointer<<endl;

return 0;
}
Code1 gives me the output as the hexadecimal value of the first
element of the array, which is what i expected the pointer to contain
(memory address), but code2 gives me the output 'words', instead of
the hexadecimal value of the first element of the arrray.

Am i missing something out in understanding the concept?

Will appreciate some help on explaining that.

Thanks,

TJ

[David Hilsee]

"ali" <[e-mail]> wrote in message
news:94********************************@4ax.com...

Hi,

I'm trying to understand the reason for different output on the
following codes

Code1:

#include <iostream.h>

int main()
{
int array[]={2,4,5};

int *pointer =0;

pointer=array;

cout<<pointer<<endl;

return 0;
}
Code2:

#include <iostream.h>

int main()
{
char array[]="words";

char *pointer =0;

pointer=array;

cout<<pointer<<endl;

return 0;
}
Code1 gives me the output as the hexadecimal value of the first
element of the array, which is what i expected the pointer to contain
(memory address), but code2 gives me the output 'words', instead of
the hexadecimal value of the first element of the arrray.

Am i missing something out in understanding the concept?

Will appreciate some help on explaining that.

In C (and also in C++, because it came from C), character arrays with a null
byte at the end can be used as strings. In C++, we have other standard
options for manipulating characters, but in C, the plain old null-terminated
array is it. Thanks to the existence of the C-style string, a lot of code
that takes a char */const char */etc assumes that the pointer is a pointer
to the first element of a null-terminated character array that should be
treated as a string. You have just stumbled upon one of those methods. The
array that you created is a null-terminated array of chars of size 6 (+1 for
the terminator), and you passed a pointer to the first element to a method
that takes a pointer to an array of characters and assumes that you wish to
treat them as a C-style string. Therefore, your code does not display the
address and instead displays the contents of the array.

--
David Hilsee

[John Harrison]

On Tue, 3 Aug 2004 21:20:56 -0400, David Hilsee
<da*************@yahoo.com> wrote:

"ali" <[e-mail]> wrote in message
news:94********************************@4ax.com...

Hi,

I'm trying to understand the reason for different output on the
following codes

Code1:

#include <iostream.h>

int main()
{
int array[]={2,4,5};

int *pointer =0;

pointer=array;

cout<<pointer<<endl;

return 0;
}
Code2:

#include <iostream.h>

int main()
{
char array[]="words";

char *pointer =0;

pointer=array;

cout<<pointer<<endl;

return 0;
}
Code1 gives me the output as the hexadecimal value of the first
element of the array, which is what i expected the pointer to contain
(memory address), but code2 gives me the output 'words', instead of
the hexadecimal value of the first element of the arrray.

Am i missing something out in understanding the concept?

Will appreciate some help on explaining that.

In C (and also in C++, because it came from C), character arrays with a
null
byte at the end can be used as strings. In C++, we have other standard
options for manipulating characters, but in C, the plain old
null-terminated
array is it. Thanks to the existence of the C-style string, a lot of
code
that takes a char */const char */etc assumes that the pointer is a
pointer
to the first element of a null-terminated character array that should be
treated as a string. You have just stumbled upon one of those methods.
The
array that you created is a null-terminated array of chars of size 6 (+1
for
the terminator), and you passed a pointer to the first element to a
method
that takes a pointer to an array of characters and assumes that you wish
to
treat them as a C-style string. Therefore, your code does not display
the
address and instead displays the contents of the array.

And should you wish to display the address instead of the content of the
string, change your code like this

cout << static_cast<void*>(pointer) << endl;

By casting your pointer from char* to void* you ensure that it is treated
as a memory address not as a pointer to a string.

john

[Method Man]

"David Hilsee" <da*************@yahoo.com> wrote in message
news:DM********************@comcast.com...

"ali" <[e-mail]> wrote in message
news:94********************************@4ax.com...

Hi,

I'm trying to understand the reason for different output on the
following codes

Code1:

#include <iostream.h>

int main()
{
int array[]={2,4,5};

int *pointer =0;

pointer=array;

cout<<pointer<<endl;

return 0;
}
Code2:

#include <iostream.h>

int main()
{
char array[]="words";

char *pointer =0;

pointer=array;

cout<<pointer<<endl;

return 0;
}
Code1 gives me the output as the hexadecimal value of the first
element of the array, which is what i expected the pointer to contain
(memory address), but code2 gives me the output 'words', instead of
the hexadecimal value of the first element of the arrray.

Am i missing something out in understanding the concept?

Will appreciate some help on explaining that.
In C (and also in C++, because it came from C), character arrays with a

null byte at the end can be used as strings. In C++, we have other standard
options for manipulating characters, but in C, the plain old null-terminated array is it. Thanks to the existence of the C-style string, a lot of code
that takes a char */const char */etc assumes that the pointer is a pointer
to the first element of a null-terminated character array that should be
treated as a string. You have just stumbled upon one of those methods. The array that you created is a null-terminated array of chars of size 6 (+1 for the terminator), and you passed a pointer to the first element to a method
that takes a pointer to an array of characters and assumes that you wish to treat them as a C-style string. Therefore, your code does not display the
address and instead displays the contents of the array.

--
David Hilsee

Man, that's confusing for n00bs. I mean, if he really wanted to print the
contents of the string, a simple pointer dereference would suffice.

You'd think they would have modified the code for C-style strings.

[John Harrison]

On Wed, 4 Aug 2004 02:07:41 -0400, Method Man <a@b.c> wrote:

"David Hilsee" <da*************@yahoo.com> wrote in message
news:DM********************@comcast.com...

"ali" <[e-mail]> wrote in message
news:94********************************@4ax.com...

> Hi,
>
> I'm trying to understand the reason for different output on the
> following codes
>
> Code1:
>
> #include <iostream.h>
>
> int main()
> {
> int array[]={2,4,5};
>
> int *pointer =0;
>
> pointer=array;
>
> cout<<pointer<<endl;
>
> return 0;
> }
>
>
> Code2:
>
> #include <iostream.h>
>
> int main()
> {
> char array[]="words";
>
> char *pointer =0;
>
> pointer=array;
>
> cout<<pointer<<endl;
>
> return 0;
> }
>
>
> Code1 gives me the output as the hexadecimal value of the first
> element of the array, which is what i expected the pointer to contain
> (memory address), but code2 gives me the output 'words', instead of
> the hexadecimal value of the first element of the arrray.
>
> Am i missing something out in understanding the concept?
>
> Will appreciate some help on explaining that.
In C (and also in C++, because it came from C), character arrays with a

null

byte at the end can be used as strings. In C++, we have other standard
options for manipulating characters, but in C, the plain old

null-terminated

array is it. Thanks to the existence of the C-style string, a lot of
code
that takes a char */const char */etc assumes that the pointer is a
pointer
to the first element of a null-terminated character array that should be
treated as a string. You have just stumbled upon one of those methods.

The

array that you created is a null-terminated array of chars of size 6 (+1

for

the terminator), and you passed a pointer to the first element to a
method
that takes a pointer to an array of characters and assumes that you wish

to

treat them as a C-style string. Therefore, your code does not display
the
address and instead displays the contents of the array.

--
David Hilsee

Man, that's confusing for n00bs. I mean, if he really wanted to print the
contents of the string, a simple pointer dereference would suffice.

It is confusing but a pionter dereference would have printed the character
that the pointer was pointing at, not the whole string. That behaviour is
at least consistent, irrespective of the type of the pointer.
You'd think they would have modified the code for C-style strings.

What would you expect the following to print?

cout << "hello world\n";

"hello world\n", is a character array just like ali's example, and
obviously you wouldn't want the address of that array printed you want the
string printed.

john

[Julián Albo]

Method Man wrote:

Man, that's confusing for n00bs. I mean, if he really wanted to print the
contents of the string, a simple pointer dereference would suffice.

If you dereference a pointer to char you obtain a char, not a string.

And yes, can be confusing, many things in C++ can be. Simply because not be
confusing to newbies is not between the design principles of C++. It will
be impossible to do that and be highly compatible with C at the same time.

--
Salu2