This

Quote:

---

The keyword this represents within a class the address in memory of the object of that class that is being executed. It is a pointer whose value is always the address of the object.
It can be used to check if a parameter passed to a member function of an object is the object itself.

---

Taken from the cplusplus.com tutorial... I dont really understand what this sentence means...they give this example...i dont understand what's special about it...

Code:

---

//this
#include <iostream.h>

class CDummy
{
public:
        int isitme(CDummy& param);
};

int CDummy::isitme(CDummy& param)
{
        if(&param == this) return 1;
        else return 0;
}

void main()
{
        CDummy a;
        CDummy *b = &a;

        if(b->isitme(a))
                cout <<"yep, &a is b\n";
        else
                cout << "nope, don't know this one\n";
}

---

i am not sure what the this pointer is...
Thanks in advance!

Code:

---

//this
#include <iostream.h>

class CDummy
{
public:
        int isitme(CDummy& param);
};

int CDummy::isitme(CDummy& param)
{
        if(&param == this) return 1;
        else return 0;
}

void main()
{
        CDummy a;
        CDummy *b = &a;

        if(b->isitme(a))
                cout <<"yep, &a is b\n";
        else
                cout << "nope, don't know this one\n";
}

---

What I dont get here, is what does the this pointer contain when i call the function: the address of *b or a?

i dont knoe why, but the &param shows up as¶m =)

Hmm bizarre thing there :)

Anyway, this is a pointer to the instance of that class. What you're doing with the second one is you're passing a pointer rather than a class, so it shouldn't even compile :confused:

Code:

---

#include <iostream>

using namespace std; // Just making you standards-compliant :)

class CDummy {
public:
    int isitme(CDummy *param); // Using pointers instead, it's easier
};

int CDummy::isitme(CDummy *param) {
    if(param == this) return 1;
    return 0;
}

int main() {
    CDummy a;
    CDummy *b = &a;

    if(b->isitme(&a))
        cout <<"yep, &a is b\n";
    else
        cout << "nope, don't know this one\n";

    return 0;
}

---

That should print "yep, &a is b" :)

he probably typed &param but it got &param
param is a reference, didn't you notice parksie? :p it should compile

Code:

---

CDummy a;
CDummy *b = &a;

if(b->isitme(a))

---

It's expecting a reference not a pointer :confused: Wouldn't passing a pointer violate the rules of referencing?

Eh, well a is not a pointer right? if b is null (for some odd reason), it still fires isitme and pass a right?

Ok I can't read...I thought b was a :rolleyes:

hehehe :p

I really need to get a good monitor :eek:

Thanks you very much guys!!