CODEWITHSUNDEEP
c++constructor
Terry Li
Terry Li

Reputation: 17268

C++ constructor confusion

MyObject obj = new MyObject(para1);

In Java I can always do this while switching to C++ it gives me an error:conversion from ‘MyObject*’ to non-scalar type ‘MyObject’ requested.

It seems the way to fix it is MyObject* obj = new MyObject(para1);.

What if I want the object itself rather than the pointer to the object?

Would MyObject obj(para1);be the only way that works for me?

Since I'm switching from Java to C++, I also wonder:

MyObject* obj1 = new MyObject(para1);

If I print out the value of obj1, it would be an address.

MyObject obj2(para1);

What would be the printed-out value for obj2?

Upvotes: 1

Views: 108

Answers (6)

Loki Astari
Loki Astari

Reputation: 264649

It seems the way to fix it is MyObject* obj = new MyObject(para1);.

Yes. The result of new is always a pointer to an object.
Note in C++ it is very rare to use RAW pointers (you normally put a pointer into a smart pointer so that it will get correctly deleted).

What if I want the object itself rather than the pointer to the object?

MyObject obj(para1)   // Creates an normal automatic object
                      // The destructor for this obect is automatically called
                      // When the object goes out of scope (even when exceptions are thrown)

I also wonder:

MyObject* obj1 = new MyObject(para1);

If I print out the value of obj1, it would be an address.

If you print this out:

std::cout << obj1;  // print the address (normally).

std::cout << (*obj1); // Will print the object/
                      // Assuming you have defined the output operator.

MyObject obj2(para1);
std::cout << obj2;    // Again prints the object
                      // Assuming you have defined the output operator.
std::cout << &obj2;   // prints the address.

Upvotes: 0

Benjamin Lindley
Benjamin Lindley

Reputation: 103741

Would MyObject obj(para1);be the only way that works for me?

Yes, that is the normal way.

What would be the printed-out value for obj2?

If you haven't overloaded the correct operator, you would get a compiler error. You define how your object is printed, using a function with the following signature:

std::ostream & operator<<(std::ostream & os, const MyObject & obj);

Upvotes: 4

StevieG
StevieG

Reputation: 8729

You've pretty much answered your own question. If you don't want a pointer then don't use new.. If you're new to c++ and memory management, and you really do need to use a pointer, I'd strongly suggest using a smart pointer (e.g. boost::shared_ptr) instead of a raw pointer..

As for the second question, it depends on what you mean by print out. You can't just print out an object, but you can call a method that knows how to print that object out. You'd have to define this method of course. Something like this:

std::ostream& operator<<(std::ostream& os, const MyObject& rhs)

Upvotes: 0

Omtara
Omtara

Reputation: 3001

In C++, the new operator returns a pointer to the object. To get the object, you need to dereference the pointer using the * operator (e.g. *obj).

Upvotes: 0

EGOrecords
EGOrecords

Reputation: 1969

Think about the Pointer-Operator -> or (*obj).func()

Upvotes: -1

Oliver Charlesworth
Oliver Charlesworth

Reputation: 272687

In some cases*, you can also do:

MyObject obj = MyObject(para1);

Unlike in Java, objects don't inherit a toString method from some base class. So you can't print out the value of obj2 unless you've defined a mechanism to do so. The standard way is to overload operator<<:

class MyObject
{
public:
    int x;
};

std::ostream &operator<<(std::ostream &os, const MyObject &obj)
{
    os << obj.x;
    return os;
}

...

MyObject obj;
obj.x = 42;
std::cout << obj << "\n";

* Namely, if the class has a public copy constructor.

Upvotes: 3

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