How to correctly forward/wrap static_cast? - c++

Question

Basicly, I need a static_cast function wrapper to use as a predicate (for conversion), since static_cast directly is not usable in this way. Lambda is not preferred this time. My implemention:
template<typename T>
struct static_cast_forward{
template<class U>
T operator()(U&& u) const {
return static_cast<T>(std::forward<U>(u));
}
};
First, while I do have a fundamental understanding of rvalue references, etc. I want to verify if this is the correct way to implement this forward/wrapper ?
Second, to ask if any std or boost library already provide this functionality ?
Extra: would you forward other casts in the same way ?
ACTUAL CASE:
My actual case is to use with boost::range, something like:
//auto targetsRange = mixedRange | boost::adaptors::filtered(TYPECHECK_PRED) | boost::adaptors::transformed(static_cast_forward<TARGET_PTR_TYPE>());
WORKING EXAMPLES:
#include <algorithm>
template<typename T>
struct static_cast_forward {
template<class U>
T operator()(U&& u) const {
return static_cast<T>(std::forward<U>(u));
}
};
//example 1:
void test1() {
std::vector<double> doubleVec{1.1, 1.2};
std::vector<int> intVec;
std::copy(doubleVec.begin(), doubleVec.end(), intVec.end());//not ok (compiles, but gives warning)
std::transform(doubleVec.begin(), doubleVec.end(), std::back_inserter(intVec), static_cast_forward<int>()); //ok
}
//example 2:
struct A {
virtual ~A() {}
};
struct B : public A {
};
struct C : public A {
};
void test2() {
std::vector<A*> vecOfA{ new B, new B};
std::vector<B*> vecOfB;
//std::transform(vecOfA.begin(), vecOfA.end(), std::back_inserter(vecOfB), static_cast<B*>); //not ok: syntax error..
std::transform(vecOfA.begin(), vecOfA.end(), std::back_inserter(vecOfB), static_cast_forward<B*>() ); //ok
}

Answer 1

Addition after the question has been clarified.
In both cases you don't need to std::forward anything, because there is nothing to move, you need just a cast. But if you want to generalize to movable types, too, then your implementation seems fine to me. Just don't call it forward because it is not a forward. As far as I know, there is nothing in std to mimic your struct.
So I would just add test3() that really needs moving:
struct B { };
struct D {
explicit D(B&&) { } // Note: explicit!
};
void test3()
{
std::vector<B> vb{B{}, B{}};
std::vector<D> vd;
// Won't compile because constructor is explicit
//std::copy(std::make_move_iterator(vb.begin()), std::make_move_iterator(vb.end()), std::back_inserter(vd));
// Works fine
std::transform(std::make_move_iterator(vb.begin()), std::make_move_iterator(vb.end()), std::back_inserter(vd), static_cast_forward<D>());
}
Answer before question was clarified.
If I correctly understood your intention, then this is what you want:
template<typename T>
struct static_cast_forward {
template<class U>
decltype(auto) operator()(U&& u) const
{
if constexpr (std::is_lvalue_reference_v<U>)
return static_cast<T&>(u);
else
return static_cast<T&&>(u);
}
};
Then you have:
struct B { };
struct D : B { };
void foo() {
D d;
static_cast_forward<B> scf_as_B;
static_assert(std::is_same_v<decltype(scf_as_B(D{})), B&&>);
static_assert(std::is_same_v<decltype(scf_as_B(d)), B&>);
}