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.
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.
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.
