The function f actually takes a function pointer to a function that takes no arguments and gives a Bar. The type of the argument to f is Bar (*)().
This code fails to compile (and we can see the actual type of the argument in the error message):
class Foo { };
class Bar { };
Foo f(Bar());
int main() {
Bar b;
f(b);
return 0;
}
But this code does compile:
class Foo { };
class Bar { };
Foo f(Bar());
Bar g();
int main() {
f(g);
return 0;
}
The second meaning it could have, as you say in the question, is that you are making a new Foo object called f and you are calling the constructor with Bar() (a new instance of Bar). It would be similar to:
Foo f = Foo(Bar());
In this situation of Foo f(Bar()); though, the first interpretation is chosen by the compiler.
Somewhat confusingly, if you add another set of parentheses, as in
Foo f((Bar()));
the compiler picks the second interpretation.
