[英]Bound member-function pointer
I was wondering why there is no concept like "bound member-function pointer", ie a member function pointer that is bound to a specific object and can therefore can be treated like a normal function pointer, eg我想知道为什么没有像“绑定成员函数指针”这样的概念,即绑定到特定对象的成员函数指针,因此可以像普通函数指针一样对待,例如
#include <functional>
struct Foo {
void foo() {}
};
void foo() {}
void call(void(*)()) {}
template<typename T>
void call_mem(void(T::*fn)(), T* x) { (x->*fn)(); }
int main() {
Foo x{};
call(&foo); // works fine
call_mem<Foo>(&Foo::foo, &x); // okay, why though
call(std::bind(std::mem_fn(&Foo::foo), &foo)); // doesnt work and looks horrible + return type of mem_fn is unspecified
call(std::bind(std::function(&Foo::foo), &foo)); // doesnt work
// something like this?
call(&x::foo); // imaginary
call(&Foo::foo(x)); // imaginary
}
the best I could come up with is this我能想到的最好的就是这个
#include <memory>
void call(Function<void> *fn) { (*fn)(); }
struct Foo {
void foo() {}
};
void foo() {}
int main(int argc, char **argv) {
auto x = [] {};
Foo f{};
std::unique_ptr<Function<void>> fn0{new Function{&foo}};
std::unique_ptr<Function<void>> fn1{new MemberFunction{&f, &Foo::foo}};
std::unique_ptr<Function<void>> fn2{new MemberFunction{x}};
call(fn0.get());
call(fn1.get());
call(fn2.get());
}
using this utility class使用这个实用程序类
#include <type_traits>
#include <utility>
template <typename R, typename... Args>
struct Function {
using Fn = R (*)(Args...);
Fn fn;
explicit Function(Fn fn) : fn{fn} {}
virtual R operator()(Args... args) noexcept(
noexcept(fn(std::forward<Args>(args)...))) {
return fn(std::forward<Args>(args)...);
}
};
template <typename R, typename... Args>
Function(R (*)(Args...)) -> Function<R, Args...>;
template <typename T, typename = std::void_t<>>
struct MemberFunction;
template <typename T, typename R, typename... Args>
struct MemberFunction<R (T::*)(Args...)> final : Function<R, Args...> {
using Fn = R (T::*)(Args...);
T *obj;
Fn fn;
MemberFunction(T *obj, Fn fn)
: Function<R, Args...>{nullptr}, obj{obj}, fn{fn} {}
R operator()(Args... args) noexcept(
noexcept((obj->*fn)(std::forward<Args>(args)...))) override {
return (obj->*fn)(std::forward<Args>(args)...);
}
};
template <typename T, typename R, typename... Args>
struct MemberFunction<R (T::*)(Args...) const> : Function<R, Args...> {
using Fn = R (T::*)(Args...) const;
const T *obj;
Fn fn;
MemberFunction(const T *obj, Fn fn)
: Function<R, Args...>{nullptr}, obj{obj}, fn{fn} {}
R operator()(Args... args) noexcept(
noexcept((obj->*fn)(std::forward<Args>(args)...))) final override {
return (obj->*fn)(std::forward<Args>(args)...);
}
};
template <typename T>
struct MemberFunction<T, std::void_t<decltype(&T::operator())>> final
: MemberFunction<decltype(&T::operator())> {
explicit MemberFunction(const T &obj)
: MemberFunction<decltype(&T::operator())>{&obj, &T::operator()} {}
};
template <typename T>
MemberFunction(T) -> MemberFunction<T>;
template <typename T, typename R, typename... Args>
MemberFunction(T *, R (T::*)(Args...)) -> MemberFunction<R (T::*)(Args...)>;
template <typename T, typename R, typename... Args>
MemberFunction(const T *, R (T::*)(Args...) const)
-> MemberFunction<R (T::*)(Args...) const>;
Yes.是的。 Use lambda's
使用 lambda 的
class Foo {
public:
void Bar() const {}
};
#include <functional>
void Baz(std::function<void()> const& fun) {
fun();
}
int main() {
Foo foo;
std::function<void()> fooBar = [&foo]() { foo.Bar(); };
Baz(fooBar);
}
edit: more reading here编辑:更多阅读在这里
edit 2: using std::bind
for this has massive overhead... Just look at this .编辑 2:为此使用
std::bind
有巨大的开销......看看这个。 Lambda's are practically zero cost. Lambda 的成本几乎为零。
Yes ,是的,
you could use Lambda expressions, that from c++11 have replaced all these binding problem.你可以使用Lambda表达式,从 c++11 开始已经取代了所有这些绑定问题。
See this link for more details: Lambda Expression有关更多详细信息,请参阅此链接: Lambda 表达式
Remember that from C++14 the lambda expressions are also polymorphic.请记住,从 C++14 开始,lambda 表达式也是多态的。
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