awk 'NR==FNR{a[$1]=$2;next} a[$3]{print $0 " " a[$3]}' cc4 bed_chr_4.bed > outfile
SpecificDerived 是否可以引用正在使用的特定Derived类?这样的
awk 'NR==FNR{a[$1]++;b[$1]=$2;next} a[$3] {print $0 " " b[$3]}' cc4 bed_chr_4.bed > outfile
会实例化从Derived1继承的BaseDecorator吗?
之所以出现问题,是因为我需要为库类及其所有派生类提供一个装饰器,但希望保持代码尽可能干燥。
谢谢!
答案 0 :(得分:2)
如果我正确理解了您的问题,则希望您的BaseDecorator继承自特定的Derived类。
在这种情况下,您可以执行以下操作:
#include <iostream>
#include <type_traits>
class Base {
public:
virtual void f1() {
std::cout << "Base::f1" << std::endl;
}
};
class Derived1 : public Base {
public:
void f1() override {
std::cout << "Derived1::f1" << std::endl;
}
};
class Derived2 : public Base {
public:
void f1() override {
std::cout << "Derived2::f1" << std::endl;
}
};
class Derived3 {
public:
void f1() {
std::cout << "Derived3::f1" << std::endl;
}
};
template <typename T,
typename = typename std::enable_if<std::is_base_of<Base, T>::value>::type >
class BaseDecorator;
template <typename T>
class BaseDecorator<T>: public T {
public:
void f2() {
T::f1();
}
};
int main() {
BaseDecorator<Derived1> bd1;
bd1.f2();
BaseDecorator<Derived2> bd2;
bd2.f2();
//BaseDecorator<Derived3> bd3; // Compilation fails !!!
//bd3.f2(); // Compilation fails !!!
return 0;
}
输出:
Derived1::f1
Derived1::f2