On Wed, 4 Aug 2004 02:07:41 -0400, Method Man <a@b.c> wrote:
[color=blue]
>
> "David Hilsee" <davidhilseenews@yahoo.com> wrote in message
> news:DMidncSCLZ15pI3cRVn-qQ@comcast.com...[color=green]
>>
>> "ali" <tj@raha.com> wrote in message
>> news:94d0h0tacu9hr3soc4jnqah6bvj4ko2io0@4ax.com...[color=darkred]
>> > 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.[/color]
>>
>> In C (and also in C++, because it came from C), character arrays with a[/color]
> null[color=green]
>> 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[/color]
> null-terminated[color=green]
>> 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.[/color]
> The[color=green]
>> array that you created is a null-terminated array of chars of size 6 (+1[/color]
> for[color=green]
>> 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[/color]
> to[color=green]
>> 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
>>
>>[/color]
>
> 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.
>[/color]
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.
[color=blue]
> You'd think they would have modified the code for C-style strings.
>[/color]
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
