Why we are not using a pointer in instantiating the entity object in the code given below of C++?

Question

In the main function in this code what does this (ScopedPtr ent = new Entity()) mean Why we are not using (ScopedPtr*) as per C++ instantiating style

#include<string>

class Entity
{
public:
    Entity()
    {
        std::cout << "Created Entity!" << std::endl;
    }
    void Print()
    {
        std::cout << "Print" << std::endl;
    }
    
};

class ScopedPtr
{
private:
    Entity* m_Ptr;
public:
    ScopedPtr( Entity* ptr) 
        : m_Ptr(ptr)
    {}

    /*ScopedPtr()
    {   
        std::cout << "Hello"; 
    }*/

    ~ScopedPtr()
    {
        std::cout << "Deleted Pointer";
        delete m_Ptr;
    }
};

int main()
{
    {
        ScopedPtr ent = new Entity();

And why the ScopedPtr(Entity Constructor) Didn't take a Entity* parameter and the code ran successfully

    }
    std::cin.get();
}

the constructor is ScopedPtr( Entity* ptr). It does have an Entity* parameter — M.M
I am Talking about ScopedPtr ent = new Entity() in this code it didn't take a actual parameter in the main function. — Ishit Jaiswal

JDługosz JDługosz · Accepted Answer · 2021-04-22T14:09:53

ScopedPtr ent = new Entity();

The right-hand-side is a value of type Entity*, regardless of the details. So it could be written as

Entity* t = new Entity();  // doesn't matter what it's = to, just note its type.
ScopedPtr ent = t;

Now you have a variable definition T x = y; where y is not of type T. So, it implicitly converts to the correct type, and this is really

ScopedPtr ent = ScopedPtr(t);

The = here is initialization not assignment, so ent is initialized with that (converted) right-hand-side.

Why we are not using a pointer in instantiating the entity object in the code given below of C++?

2 Answers