c++ - About conversion operator and operator() -

in following code:

template <class t> class mval { protected:     t max; public:     template <class iter> mval (iter begin, iter end):max(*begin) {         while(begin != end) {             (*this)(*begin);             ++begin;         }     }     void operator()(const t &t) {         if (t > max)             max = t;     }     void print() {         cout << max;     } };  int main() {     int arr[3] = { 10,20,5 };     (mval<int>(arr, arr + 3)).print(); } 

why (*this)(*begin); leads operator()?

shouldn't go operator when have mval x; x(t); ? behaves it's conversion operator, how?

how mval x; x(t); different (*this)(*begin);? in both cases see object of type mval followed parentheses 1 argument inside. did expect happen? (*this) not type, it's lvalue of type mval, don't see how "behaves it's conversion operator" either.