策略模式的实例化通常被示例完全忽略。假设有一个输入定义要使用哪个类。我们会得到一些类似的东西:
class Strategy {
Strategy(){}
virtual void runAlgorithm() = 0;
};
class A : public Strategy {
A () {}
static bool isA(char input){ return input == 'A'; }
void runAlgorithm { /* Do A algorithm */ }
};
class B : public Strategy {
B () {}
static bool isB(char input){ return input == 'B'; }
void runAlgorithm { /* Do B algorithm */ }
};
// Other algorithms
Strategy* createStrat(char input){
Strategy* instance;
// Define which algorithm to use
if (A::isA(input)) {
instance = A();
} else if (B::isB(input)) {
instance = B();
} ...
// Run algorithm
instance.runAlgorithm();
return instance;
}
正如你所看到的，如果我们有多种不同的算法，这个 if/switch 可能会变得相当大。是否有一种模式可以使该代码更易于人为解析(即 for 循环和调用)，而无需添加对数组？这个问题也可以扩展到“如何实例化基于输入的策略模式？”
不要限制于此代码，因为它只是一个示例。
最佳答案
好吧，如果您提前知道所有策略，您可以使用非常简单的元编程递归来自动展开 if-else 链。我们开始吧:
#include <string_view>
#include <iostream>
#include <exception>
#include <memory>
struct Strategy {
Strategy(){}
virtual void runAlgorithm() = 0;
};
struct A : public Strategy {
A () {std::cout << "Creating A" << std::endl;}
static constexpr std::string_view id(){
return std::string_view("A");
}
void runAlgorithm() { /* Do A algorithm */ }
};
struct B : public Strategy {
B () {std::cout << "Creating B" << std::endl;}
static constexpr std::string_view id(){
return std::string_view("B");
}
void runAlgorithm() { /* Do B algorithm */ }
};
struct C : public Strategy {
C () {std::cout << "Creating C" << std::endl;}
static constexpr std::string_view id(){
return std::string_view("C");
}
void runAlgorithm() { /* Do C algorithm */ }
};
// the if else chains are constructed by recursion
template <class Head, class... Tail>
struct factory {
static std::unique_ptr<Strategy> call(std::string id) {
if(Head::id() == id) return std::make_unique<Head>();
else return factory<Tail...>::call(id);
}
};
// specialization to end the recursion
// this throws an exception, but you can adapt it to your needs
template <class Head>
struct factory<Head> {
static std::unique_ptr<Strategy> call(std::string id) {
if(Head::id() == id) return std::make_unique<Head>();
else throw std::invalid_argument("The strategy id you selected was not found.");
}
};
// here is your factory which can create instances of A,B,C based only on the runtime id
using my_factory = factory<A,B,C>;
int main() {
auto Astrategy = my_factory::call("A");
auto Bstrategy = my_factory::call("B");
auto Cstrategy = my_factory::call("C");
my_factory::call("D"); // exception thrown
}
实时代码 here
编辑
按照 Jarod42 的建议进行编辑以考虑智能指针和错误检查。
关于c++ - 如何实例化基于输入的策略模式，我们在Stack Overflow上找到一个类似的问题： https://stackoverflow.com/questions/51493703/