如何在 C++ 中从基 class 和 return 派生 class 继承后记增量

Question

我有几个具有类似行为的 c++ classes。此外，大多数 class 方法都可以从几个基本方法中构建出来。所以我想用派生方法定义一个基础class，从基础class继承并在派生class中定义剩余的方法。这是我尝试使用 CRTP

template <class derived_class> class base_class {
public:
    virtual derived_class& operator++ () = 0;
    virtual derived_class& fun1() = 0;
    derived_class operator++ (int) {
        derived_class toreturn(static_cast<derived_class&>(*this));
        ++*this;
        return toreturn;}
    derived_class& fun2() {
        this->fun1();
        return static_cast<derived_class&>(*this);
    };
};

class deriv1 : public base_class<deriv1> {
public:
   int n;
    deriv1():n(0){};
    deriv1(deriv1& other):n(other.n){};
    deriv1& operator++ () override { ++n; return *this;}
    deriv1& fun1() override { n *= n; return *this;}
};

我不明白为什么 fun2() 有效但 postscript 增量无效。如果我在派生对象上调用后记增量，我会收到错误消息“无法增加类型 'deriv1'”的值。

Answer 1

解决方法是添加一条using语句：

class deriv1 : public base_class<deriv1> {
public:
    ....
    using base_class::operator++; 
};

问题是函数解析失败。让我们想一个更简单的解决方案来说明问题：

struct Base
{
    void f()    {}
    void f(int) {}
};

struct Derived: public Base
{
    void f()    {}
};

int main()
{
    Derived a;
    a.f(1);     // This fails as there is no f() that takes an integer
                // in Derived. And since the compiler found an f() in
                // Derived it stopped looking further up the chain
                // for additional matches.
}

本题同法解决

struct Derived: public Base
{
    using Base::f;
    void f()    {}
};

如何在 C++ 中从基 class 和 return 派生 class 继承后记增量

How to inherit postscript increment from base class and return derived class in c++

c++

inheritance

post-increment