Why is this a valid function declaration in C++?

Question

Foo x(Bar());

I can write in inside another function and it compiles. Why? Also how a function declaration can get a temporary object in it's signature?

Answer 1

Well, you can do this:

Foo x();

This declares a function x that takes no arguments and returns Foo.

We can add parameters:

Foo x(int y, char z);

A parameter is just a variable declaration (x and z here).

We can omit the parameter names in a function declaration:

Foo x(int, char);

And we can declare a parameter as a function:

Foo x(Bar y());

Here we declare x as a function taking another function y (that takes no arguments and returns Bar) returning Foo.

Finally we can omit the parameter name here, too:

Foo x(Bar ());  // a function taking a function

That's how the syntax works.

Semantically, you might object, this is invalid because functions aren't value types. You can't copy a function, so you can't pass it by value. Which is true, but there's a rule that says any parameter declared as a function is silently adjusted to be a pointer by the compiler:

Foo x(Bar y());
// really means:
Foo x(Bar (*y)());

x takes a pointer to a function (taking no arguments and returning Bar).

Or without parameter names:

Foo x(Bar ());
// same as:
Foo x(Bar (*)());

This is similar to the rule that turns a parameter declared as an array into a pointer:

Foo x(Bar [42]);
// same as:
Foo x(Bar *);