Most vexing parse confusion

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Here:

auto dv = Timer();

You have an object of type Timer called dv that is being copy-initialized from a temporary (the expression on the right side of the = sign).

When using auto to declare a variable, the type of that variable is the same as the type of the expression that initializes it – not considering cv-qualifiers and references here.

In your case, the expression that initializes dv has type Timer, and so dv has type Timer.

Here:

int time_keeper(Timer());

You declare a function called time_keeper that returns an int and takes as its input a pointer to a function which returns a Timer and takes no argument.

And why isn’t the argument Timer (*) () ?

Functions decay to pointers when passed as an argument, so the type of time_keeper is actually int(Timer(*)()).

To convince yourself, you could try compiling this little program:

#include <type_traits>

struct Timer { };
int main()
{
    int time_keeper(Timer());
    static_assert(
        std::is_same<
            decltype(time_keeper), 
            int(Timer(*)())
        >::value, 
        "This should not fire!");
}

Here is a live example.

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