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C++: Is there a way to call an operator function through a static method in this context?
This is the code I have. It compiles; but I want to call the function h with the arguments of H<T>::operator(). How would I do that?
#include <utility>
#include <iostream>
template <template <class> class H, typename F>
struct N {
N() = delete;
static F f() {
H<F> h;
return h();
}
};
template <class T>
struct H {
H() = default;
T operator()() {
return T();
}
};
template <template <class> class T>
int h() {
return N<T, decltype( std::declval<T<int>>().operator()() )>::f();
}
int main() {
std::cout << h<H>();
}
For example, instead of doing h<H>() I do h<H>(5) and it outputs 5 for me through H<int>::operator(). And that also means calling the static member function f with those arguments. Can someone propose an example in which this is possible? Thanks.
1 Answer
- ?Lv 78 years agoFavorite Answer
I am not sure I follow, are you looking for something like this:
#include <utility>
#include <iostream>
template <template <class> class H, typename F>
struct N {
N() = delete;
static F f(const F& val = F()) {
H<F> h;
return h(val);
}
};
template <class T>
struct H {
H() = default;
T operator()(const T& val = T()) {
return val;
}
};
template <template <class> class T>
int h(int val = 0) {
return N<T, decltype( std::declval<T<int>>().operator()() )>::f(val);
}
int main() {
std::cout << h<H>() << h<H>(5) << '\n';
}
Or do you want to pass multiple arguments to h(), forward them to f(), and use them in the constructor of H's template parameter type you hardcoded as int? That shouldn't be a problem either, but it would be nice to have a testcase.
