C++ spooky constructor [duplicate]

Answer 1

Foo f3(); declares a function called f3, with a return type of Foo.

Answer 2

• f1 invokes the copy constructor after an explicit call, you were wrong on this one
• f2 is an explicit constructor call // you were wrong here too
• f3 declares a function
• f4 is again the copy constructor, like f1 // you're right here
• f5 would calls the default constructor // you're right here again

Answer 3

You've defined a function called f3 that returns a foo in case 3. In case 5, you have no default constructor defined, so you get an error.

Answer 4

This isn't what you think it is:

 Foo f3();

You may think this is an explicit call of the default constructor, but it's not. It's actually a declaration of a function named f3 which takes no parameters and returns a Foo by value.

That this is parsed as a function declaration rather than a constructor call is known as the Most Vexing Parse.

Answer 5

C++ has a rule that if a statement can be interpreted as a function declaration, it is interpreted in this way.

Hence the syntax Foo f3(); actually declares a function which takes no arguments and returns Foo. Work this around by writing Foo f3;, it will call the default constructor too (if there is one, of course).

Answer 6

The third line is parsed as declaring a function that takes no argument and returns a Foo.