Pointer confusion. Please explain :)

Today I've learned some concepts about Pointer in C++ and I have a confusion in this code snippet and its result.

Source Code:

Code:

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// Pointer array
#include <iostream>
using namespace std;

void main () {
        char * say = "Hello";               
        //
        cout << "say = " << say << endl;
        cout << "&say = " << &say << endl;
        cout << "*say = " << *say << endl;
        cout << "say[1]=" << say[1] << endl;
        cout << "*(say+1)=" << *(say+1) << endl;       
        //
        cout << endl << endl;
        for (int i=0; i<5; i++) {
                cout << "say["<< i <<"]=" << say[i] << endl;       
        }
        //
        cout << endl << endl;
        for (int i=0; i<5; i++) {       
                cout << "&say["<< i <<"]=" << &say[i] << endl;
        }
}

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And this is the result:

Code:

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say = Hello
&say = 0012F3A4
*say = H
say[1]=e
*(say+1)=e

say[0]=H
say[1]=e
say[2]=l
say[3]=l
say[4]=o

&say[0]=Hello
&say[1]=ello
&say[2]=llo
&say[3]=lo
&say[4]=o

---

I try to understand why it gave this result, but I cannot understand the section in bold.

Please explain it for me.

Thanks very much,
Nichya.

If say[i] is a character, then what is &say[i]? The address of the character, or char*.

char* is treated as a special case by cout to represent a C-style null-terminated string. If you want to output the address of each character, you could cast the resulting pointer to another pointer type that doesn't receive this special treatment (void*).

Code:

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for (int i=0; i<5; i++) {
    cout << "&say["<< i <<"]=" << static_cast<void*>(&say[i]) << endl;
}

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Thank you. I think I must read more to understand what you said ^^ It's a little bit complicated with me.

Quote:

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Originally Posted by nichya88

Thank you. I think I must read more to understand what you said ^^ It's a little bit complicated with me.

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Basicly, when you got "&say[0]=Hello", that's because you gave cout the address for the character 'H', and it just kept printing them all until it stumbled upon a null-termination (which is no character, the value 0, used to mark the end of the string). Ascii Table - ASCII character codes and html, octal, hex and decimal chart conversion

If say = "Hello", then say[0] would be 'H'.
That would mean that &say[0] would return the address where that character H is stored, which is the address of say[0], and also the address that say points to in the first place, since the poiner points to the first char in the string.

Thank all of you, now I have to understand it. :)

Oh and btw
SourceForge.net: Void main - cpwiki

Quote:

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Originally Posted by Elysia

Oh and btw
SourceForge.net: Void main - cpwiki

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I read some tutorials, and found that they always use int main() {} despite the fact that void main(){} is OK! . I wondered why? So, this article tells me the answer :)