Question

考虑这个输出：

Current time: 6:30 pm
Current time: 18:30
Current time: evening.
Current time: evening (for many it is dinner time, but many eat dinner later).

请注意，最后两个有一个句号，而前两个没有。我使用以下代码中的System::displayCurrentTime成员函数获得了所需的输出：

#include <iostream>
#include <string>
#include <memory>

class TimeDisplay {
    public:
        virtual std::string tell() const = 0;
        virtual std::string tellMaybeWithPeriod() const = 0;
};

class ClockDisplay12Hours : public TimeDisplay {  // #1
    std::string tell() const override {return "6:30 pm";}
    std::string tellMaybeWithPeriod() const override {return tell();}
};

class ClockDisplay24Hours : public TimeDisplay {  // #2
    std::string tell() const override {return "18:30";}
    std::string tellMaybeWithPeriod() const override {return tell();}
};

class DescriptiveTimeDisplay : public TimeDisplay {  // #3
    std::string tell() const override {return "evening";}
    std::string tellMaybeWithPeriod() const override {return tell() + ".";}
};

class CrazyDescriptiveTimeDisplay : public TimeDisplay {  // #4
    std::string tell() const override {return "evening (for many it is dinner time, but many eat dinner later)";}
    std::string tellMaybeWithPeriod() const override {return tell() + ".";}
};

struct System {
    static std::shared_ptr<TimeDisplay> timeDisplay;
    static std::string timeAsString() {return timeDisplay->tell();}
    static std::string timeAsStringMaybeWithPeriod() {return timeDisplay->tellMaybeWithPeriod();}
    // #3 and #4 will have a period, the others will not.
    static void displayCurrentTime (std::shared_ptr<TimeDisplay> t) {
        timeDisplay = t;
        std::cout << "Current time: " << System::timeAsStringMaybeWithPeriod() << '\n';
    }
    static void foo (std::shared_ptr<TimeDisplay>) {}  // #1 and #3 will have a period, the others will not.
    static void bar (std::shared_ptr<TimeDisplay>) {}  // #1, #2, and #4 will have a period, the others will not.
    static void baz (std::shared_ptr<TimeDisplay>) {}  // #2 will have a period, the others will not
};
std::shared_ptr<TimeDisplay> System::timeDisplay;

int main() {
    const std::shared_ptr<TimeDisplay> clocks[] = {std::make_shared<ClockDisplay12Hours>(), std::make_shared<ClockDisplay24Hours>(),
        std::make_shared<DescriptiveTimeDisplay>(), std::make_shared<CrazyDescriptiveTimeDisplay>()};
    for (std::shared_ptr<TimeDisplay> t : clocks)
        System::displayCurrentTime(t);
}

这不是非常混乱，但请注意，实现foo，bar，baz的下一个功能需要具有TimeDisplay的不同派生类的句点，以及实际上有超过4个这样的派生类，还有3个以上的新成员函数需要处理。是否有一种更清晰，更优雅的方式来处理这些即将到来的成员函数，而不是为每个foo，bar，baz等编写新的虚函数...期间哪个不会？以某种方式使用模板（例如重命名派生类Derived<0>，Derived<1>等......然后使用这些编译时整数来满足上述注释中规定的规则）？也许避免模板并做其他事情？

Answer 1

可以通过为四个时钟显示类提供bool模板参数来完成。 e.g。

template <bool P>
class DescriptiveTimeDisplay : public TimeDisplay {  // #3
    std::string tell() const override { return "evening"; }
    std::string tellMaybeWithPeriod() const override { return tell() + (P ? "." : ""); }
};

并控制是否通过将类作为实例来显示句点，例如

std::make_shared<DescriptiveTimeDisplay<true>>()

对于displayCurrentTime，foo，bar，baz这四个函数中的每一个，您可以通过实例化四个TimeDisplay子类来控制它们的显示格式使用不同的bool模板参数。

虚拟功能在这里和那里有所不同

1 个答案: