Creating a class object

[Jimmy Leung]

I am new to C++ and I got some problem while understanding the codes that demonstrate the creation of class objects.

I learnt that a class object (eg. class Test), can be created by

  Test A;

But how about this line

  Test A, *APtr = 0;

Is it mean that i set a Test pointer pointing to the object A i just created as well? And does the *APtr can be assigned to 0 or NULL only?

Thanks for the help.

[Ssturges]

So, the line:

Test A, *APtr = 0;

first you are creating a Test object called A (as you already knew),but
*APtr is a pointer, that can only point to Test objects.

At this very moment in the code, it is set to null (0 and null, same difference with pointers) and there is no memory allocated for this pointer;

some things you can do to APtr to allocate memory to it is:

APtr = &A; // now APtr is pointing to memory location of A; 

or

APtr = new Test(); //now you have allocated memory somewhere in your computer that can stores information for a Test object 

Now if you wanted APtr to point to the A object at declaration, this would work:

Test A, *APtr = &A;  //now APtr is pointer to the memory location of A 

is that what you were asking?

[vlad from moscow]

You could write for example

Test A, *APtr = &A;

[Jimmy Leung]

Re Ssturges:

The problems come when I test this

#include <iostream>
using namespace std;

class test{
public:
	void print(){
		cout << "Printing" << endl;
	}
};

int main(){
	test A,*Aptr = 0;
	A.print();
	Aptr -> print();
	return 0;
}

But still the line

Aptr -> print();

gives the result. Im not sure why would it be like this.

[vlad from moscow]

There is no access to memory at address 0. Simply member function print() is called and it recieves as the first argument 'this' with value 0 that is not used.

[Jimmy Leung]

Thanks very much, I think I got it.
It appears that I mixed it up with the dereference concept