Thread: what does this statement mean??

What does the last line in the following code -

Code:

b* l = new b();

do??

Code:

#include<stdio.h>
#include<iostream>

class a
{
protected:
	int i;
public:
	void fun()
	{
		std::cout<<"whatever";
	}
	a()
	{
		i = 3;
	}
};

class b: private a
{
public:
	void out()
	{
		std::cout<<i;
	}
	b(int ii)
	{
		i = ii;
	}
	b()
	{
	}
};

int main()
{
	b bb;
	b* l = new b();
}

Declares l as a pointer to an object of class b and allocates memory for it.

but why the paranthesis??

Isn't it supposed to be

Code:

b* l = new b;

??

Originally Posted by juice

but why the paranthesis??

Isn't it supposed to be

Code:

b* l = new b;

??

Same thing.
But if you wanted to call the other constructor (which takes an int), the arguments would go inside that.

Originally Posted by manasij7479

Same thing.
But if you wanted to call the other constructor (which takes an int)

???

is the statement calling a constructor?

new always calls the constructor for the target type, unless it's a native type. declaring an object on the stack as you did in line 37 also calls the constructor.

Originally Posted by Elkvis

new always calls the constructor for the target type, unless it's a native type.

am all confused man..

New would have anyways called the constructor even if we had droped the paranthesis. What purpose do they serve??

they allow you to pass a list of parameters to the constructor. they are completely optional if you are using a default constructor - one that takes no parameters.