constructor invocation mechanism

问题

struct my
{
   my(){ std::cout<<\"Default\";}
   my(const my& m){ std::cout<<\"Copy\";}
   ~my(){ std::cout<<\"Destructor\";}
};

int main()
{
   my m(); //1
   my n(my()); //2
}

Expected output :

1 ) Default
2 ) Copy

Actual output :

---

What\'s wrong with my understanding of the constructor invoking mechanism?

Note I have omitted header files for brevity.

回答1:

Case 1)

m is interpreted as a function return my and taking no arguments. To see the expected output remove () i.e use my m;

Case 2)

This is something better known as the "Most vexing parse".

n is interpreted as a function returning my that takes an argument of type pointer to function returning my taking no arguments.

To see the expected output in this case try my n((my())); [Instead of treating as an argument specification as in the former case the compiler would now interpret it as an expression because of the extra ()]

My interpretation:

my n((my())) is equivalent to my n = my(). Now the rvalue expression my() creates a temporary[i.e a call to the default constructor] and n is copy initialized to that temporary object[no call to the copy-ctor because of some compiler optimization]

P.S: I am not 100% sure about the last part of my answer. Correct me if I am wrong.

回答2:

Like Prasoon, I suspect the C++ compiler is parsing your code in a way you don't expect. For example, I think it is parsing the line

my m();

as a function prototype declaration, not as a variable declaration and call to the constructor - hence why you see no output.

来源：https://stackoverflow.com/questions/4283576/constructor-invocation-mechanism