正如Sam注意到的,MFP
是一个模板,而std::map
的第二个模板参数需要一个类型。因此,在使用函数指针填充地图之前,您需要获取实际的类型。让我提出这个案例最直接的方法 - 模板类。有了它,你将需要在对象实例化中列出所有需要的返回类型,但是你将能够使用任何类型的Call
。我将使用std::function
而不是指针,但可以轻松回滚到函数指针。
首先,我们不知道类用户需要多少类型,所以让我们将其作为可变参数。由于map
需要一个完整的类型,我们需要一堆地图 - 每种类型一个。获得它的最常见方法是一个元组,在我们的例子中需要扩展包。通过元组,我们可以在编译时搜索需要的映射,然后在运行时按名称搜索函数。看看代码与解释:
template<typename ...Types>
class B {
private:
// Template alias for std::function.
template<typename T>
using MFP = std::function<T()>;
/* Tuple of maps from std::string to MFP for all types
in Types parameter pack. */
std::tuple<std::map<std::string, MFP<Types>>...> fmap;
template<typename T>
T f() { return 2.5; }
template<typename T>
T g() { return 1.0f; }
// Call implementation with compile-time pattern matching.
// T is return type, U is current matching type
template<typename T, size_t idx, typename U, typename ...Ts>
struct CallImpl {
static T callImpl(B* this_ptr, const std::string & s) {
/* If we exhausted Ts pack, we have no proper instance for
requested return type. Let's print a human-readable
compilation error message. */
static_assert((sizeof ... (Ts)) > 0,
"Requested return type not found.");
/* Otherwise discard U, increment tuple index
and try the next type. */
return CallImpl<T, idx + 1, Ts...>::callImpl(this_ptr, s);
}
};
/* This partial specialization is called when return
* type (T in above declaration) matches
* stored type (U in above declaration). */
template<typename T, size_t idx, typename ...Ts>
struct CallImpl<T, idx, T, Ts...> {
static T callImpl(B* this_ptr, const std::string & s) {
/* First, get the map from tuple by index.
This operation is either always valid in runtime or does not compile.
Next, get function object from map. It may fail in runtime
if user passed invalid string, so consider using map::at
or add any other sensible logic for this case. */
return std::get<idx>(this_ptr->fmap)[s]();
}
};
public:
B() {
/* Populate map with objects. Ellipsis in the last line
expands Types as needed. */
fmap = std::make_tuple(std::map<std::string, MFP<Types>>{
{"f", std::bind(std::mem_fn(&B::f<Types>), this)},
{"g", std::bind(std::mem_fn(&B::g<Types>), this)}
}...);
}
template<typename T>
T Call(const std::string & s) {
/* Start pattern matching with zero index. */
return CallImpl<T, 0, Types...>::callImpl(this, s);
}
};
用法:
int main() {
B<int, float, short> a; // Provides int, float and short return types.
std::cout << a.Call<int>("f") << std::endl; // Prints 2, which is 2.5 casted to int.
std::cout << a.Call<float>("f") << std::endl; // Prints 2.5
// Compilation error with "Requested type not found." message among others.
std::cout << a.Call<double>("f") << std::endl;
}
一些注意事项:
在
2.5
f
声明是双重的文字,但双不列在B<int, float> a;
中,我们在上得到编译错误。
- 代码
Call
方法和callImpl
函数short
根本不会生成,因为我们没有实例化它。
您必须指定模板参数,如'std :: map> fmap;'。否则''A :: fmap'可能会在刚刚写入'A a;'时被实例化。 –
songyuanyao