Function Pointer declarations [duplicate] - c++

Question

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Default constructor with empty brackets
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The other day i was trying to create an object by calling the default constructor of another class, and it ended up making a function declaration, Here is an example:
struct integer {
integer(){} //Default constructor.
};
struct rational {
rational(integer n, integer d){} //Default constructor.
};
void multiply(rational(), rational()) { //Valid syntax? Takes two function pointers.
}
rational one_half() {
return rational(integer(), integer()); //Here doesnt make a function declaration.
}
int main() {
rational num(integer(), integer()); //Here makes a function declaration,
//instead of constructing a rational object.
multiply(one_half, one_half); //function taking two function pointers.
}
Why does this happen? I know and can call the constructor like this integer::integer() but i would like to understand what's happening here and why integer() is behaving like integer(*)() in this example.

Answer 1

Why does this happen?
The rule comes from C, in C++ it works the same. From c++ standard dcl.fct inside a function parameter list:
... After determining the type of each parameter, any parameter of type “array of T” or of function type T is adjusted to be “pointer to T”. ...
Inside a function parameter list a parameter that has a type of function type integer() is converted to a pointer to that type integer(*)().
So is integer() in fact equal to integer(*)()?
Inside function parameter list, yes.