C++中静态方法的局部变量范围

Question

I was trying to create an instance of an object with private constructor. Is this method correct?

#include <iostream> 
using namespace std; 
class A{
    int y;
    A(int y=2):y(y){};
    public:
    A(const A &obj){
        cout<<"copy Cotr is called"<<endl;
        this->y=obj.y;
    }
    int* addr(){
        int *a=&y;
        return a;
    }
    static A create(int y1=2){
        class A a(y1);
        cout<<&a<<endl;
        return a;
    }
    void print(){
        cout<<y<<endl;
    }
};
int main(){
    A o1=A::create(1);
    A o2=A::create(3);
    cout<<&o1<<" "<<&o2<<endl;
    return 0;
}

The output is:

0x7ffd20d2f280
0x7ffd20d2f284
0x7ffd20d2f280 0x7ffd20d2f284

In the above code, I fail to understand few points:

1. create() returns a value, so during assignment which take place in A o1=A::create(1); , copy constructor should be called but it isn't. why?
2. Is this some undefined behavior as the address of the object created in create() and in main() are same and the object in create() is local and goes out of scope after the function ends.

Answer 1

This:

A a = x();

is different than this:

A a;
a = x();

In the second case you have a copy, in the first you don't.

To answer your second question, the object created in create() is local all right, but then it's copied back to the caller (actually, moved). Since the stack frame is the same, the local address is the same. But when it's copied, you have two different objects.

Answer 2

1. create() returns a value, so during assignment which take place in A o1=A::create(1); , copy constructor should be called but it isn't. why?

You have copy-elision ( Demo ) which is an optimization allowed by standard.

The object is directly constructed in final place and no copy/move is done.

Some copy-elisions is even mandatory in C++17. (half of them in your case).

2. Is this some undefined behavior as the address of the object created in create() and in main() are same and the object in create() is local and goes out of scope after the function ends.

With copy elision, as object is directly constructed in final place, there is indeed only one address, it is as-if the local variable of create was the one from main .

Answer 3

Thanks for all the answers. Dug deep into it and now answering my own question.

For the learned:

This is major instance of copy elision in C++ 17.

Answer ends for you. For the one who still didn't understood:

This method has no errors and do not show any undefined behavior. Most of the compilers uses C++11 or C++17 these days which is well integrated with copy elision(copy omission) .

create() returns a value, so during assignment which take place in A o1=A::create(1);, copy constructor should be called but it isn't. why?

Copy constructor is not called as the return type is prvalue(pure rvalue) and the compiler optimizes the program in such a way that the return value is created directly into the storage where the function's return value would otherwise be copied or moved to. So, actually no move or copy takes place so no move/copy constructor is called. Even after the end of create() destructor too is not called.

Is this some undefined behavior as the address of the object created in create() and in main() are same and the object in create() is local and goes out of scope after the function ends.

So, even after the function create() ends, the object dont get pushed out of scope or the destructor for that object is not called. As the object is created in not in the create() but in the main() itself. It is not an undefined behavior but a major benefit of C++ as unnecessary pointers and functions need not to be declared or invoked. Here, NRVO takes place.

• In a return statement, when the operand is the name of a non-volatile object with automatic storage duration, which isn't a function parameter or a catch clause parameter, and which is of the same class type (ignoring cv-qualification) as the function return type. This variant of copy elision is known as NRVO, "named return value optimization".