Question

I read somewhere (cannot find the source now) that

MyClass *p1 = new MyClass;

and

MyClass *p2 = new MyClass();

are essentially equivalent, provided that MyClass provides a default constructor. The compiler understands what I want to do and adds the empty parentheses.

If this is the case, why I am not allowed to write

throw MyException;

but have to use

throw MyException();

? (Yep, a question mark at the beginning of a line.)

To add some more confusion, the C++ FAQ suggests that the second usecase (new MyClass()) does not invoke a constructor, but calls function defined with operator() instead.

Answer 1

The compiler understands what I want to do and adds the empty parentheses.

No it doesn’t; the two expressions aren’t quite equivalent. The difference is in how the objects are are initialised: the first uses default-initialisation, while the second uses value-initialisation. So they are equivalent for classes that define a default constructor; otherwise, the first will leave POD objects uninitialised, while the second will initialise them to zero.

why I am not allowed to write throw MyException;?

MyException() is an expression that creates a value-initialised temporary object; you can throw that just like you can throw the value of any other suitable expression.

MyException isn’t an expression; it’s just a type name. You can only throw the value of an expression, so throw MyException; is not valid. There’s no way to create a default-initialised temporary.

To add some more confusion, the C++ FAQ suggests that the second usecase (new MyClass()) does not invoke a constructor, but calls function defined with operator() instead.

No it doesn’t. It says that a declaration like List x(); declares a function with a return type List, not (as one might think) a value-initialised object of type List. It has nothing to do with new-expressions or operator().