C++ 抽象基类构造函数/析构函数 - 一般正确性

C++ abstract base class constructors/destructors - general correctness

提问by Stuart Blackler

Recently I have dumb as a developer, so I took the plunge, got a C++ book and learning how to do things properly. In my head, I know what I would like to do. I effectively want an Interfacethat when inherited, must be overridden (if this is possible?). So far, I have the following:

最近我作为一名开发人员很笨，所以我冒险尝试，买了一本 C++ 书并学习如何正确地做事。在我的脑海中，我知道我想做什么。我实际上想要一个Interface继承时，必须被覆盖（如果这可能？）。到目前为止，我有以下几点：

class ICommand{

public:
    //  Virtual constructor. Needs to take a name as parameter
    //virtual ICommand(char*) =0;
    //  Virtual destructor, prevents memory leaks by forcing clean up on derived classes?
    //virtual ~ICommand() =0; 
    virtual void CallMe() =0;
    virtual void CallMe2() =0;
};

class MyCommand : public ICommand
{
public:
    // Is this correct?
    MyCommand(char* Name) { /* do stuff */ }
    virtual void CallMe() {}
    virtual void CallMe2() {}
};

I have purposely left how I think the constructor/destructor's should be implemented in ICommand. I know if I remove the comments, it will not compile. Please could someone:

我故意离开了我认为构造函数/析构函数应该在ICommand. 我知道如果我删除评论，它将无法编译。请有人：

1. Show me how to declare the constructor/destructor's in ICommandand how they are meant to be used in MyCommand
2. Have I set things up correctly in ICommandso that MyCommandmust override CallMeand CallMe2.

1. 告诉我如何声明构造函数/析构函数的 inICommand以及它们的用途MyCommand
2. 我是否正确设置了内容，ICommand以便MyCommand必须覆盖CallMe和CallMe2.

I hope I haven't missed something really simple...

我希望我没有错过一些非常简单的东西......

回答by e.James

C++ does not allow for virtual constructors. A simple implementation (without the virtual constructor) would look something like this:

C++ 不允许使用虚拟构造函数。一个简单的实现（没有虚拟构造函数）看起来像这样：

class ICommand {
public:
    virtual ~ICommand() = 0;
    virtual void callMe() = 0;
    virtual void callMe2() = 0;
};

ICommand::~ICommand() { } // all destructors must exist

Note that even a pure virtual destructor mustbe defined.

请注意，即使是纯虚拟析构函数也必须定义。

A concrete implementation would look exactly like your example:

一个具体的实现看起来和你的例子完全一样：

class MyCommand : public ICommand {
public:
    virtual void callMe() { }
    virtual void callMe2() { }
};

You have a couple of optionsfor the constructor. One option is to disable the default constructor for ICommand, so that subclasses will haveto implement a constructor that calls your ICommand constructor:

构造函数有几个选项。一种选择是禁用 的默认构造函数ICommand，以便子类必须实现一个调用 ICommand 构造函数的构造函数：

#include <string>

class ICommand {
private:
    const std::string name;
    ICommand();
public:
    ICommand(const std::string& name) : name(name) { }
    virtual ~ICommand() = 0;
    virtual void callMe() = 0;
    virtual void callMe2() = 0;
};

ICommand::~ICommand() { } // all destructors must exist

A concrete implementation would now look something like this:

一个具体的实现现在看起来像这样：

class MyCommand : public ICommand {
public:
    MyCommand(const std::string& name) : ICommand(name) { }
    virtual void callMe() { }
    virtual void callMe2() { }
};

回答by Theo

I know this one is old, but it is still my first hit on this issue. This is how I would do it.

我知道这是旧的，但它仍然是我在这个问题上的第一次打击。这就是我要做的。

Interface header foo.h:

接口头文件 foo.h：

#pragma once
#include <memory>

enum class Implementations {Simple, Fancy};

class Foo
{
public:
    using Ptr = std::unique_ptr<Foo>;
    virtual ~Foo() = default;
    virtual void do_it() = 0;
};

Foo::Ptr create_foo(Implementations impl); // factory

Yes I know that "pragma once" is strictly speaking not standard, but it works for me.

是的，我知道“pragma once”严格来说不是标准的，但它对我有用。

Note that nothing is implemented here. There is no constructor: an abstract class can not be instantiated. You get a pointer to the interface through the factory. For the virtual function calls to work, they must be called through a pointer. The virtual destructor is defaulted because it doesn't have to do anything special except polymorphing to the implementation. The factory is a free function. No need to try to make it a static member or something like that. This is not java.

请注意，这里没有实现任何内容。没有构造函数：抽象类不能被实例化。您可以通过工厂获得一个指向接口的指针。要使虚函数调用起作用，必须通过指针调用它们。虚拟析构函数是默认的，因为它不需要做任何特殊的事情，除了对实现进行多态。工厂是一个免费的功能。无需尝试使其成为静态成员或类似的东西。这不是java。

Interface foo.cpp:

接口 foo.cpp：

#include "foo.h"
#include "foo_impl.h"

Foo::Ptr create_foo(Implementations impl)
{
    switch (impl)
    {
    case Implementations::Simple:
        return std::make_unique<Simple_foo>();
    case Implementations::Fancy:
        return std::make_unique<Fancy_foo>();
    default:
        return nullptr;
    }
}

Here the factory is implemented. Notice that the factory has to know the implementation(s). That is why we don't implement it inline: if it was inline, the interface header would have to include the implementation header, and through it, knowledge of the implementation would "leak out" to the callsite.

这里实现了工厂。请注意，工厂必须知道实现。这就是我们不内联实现它的原因：如果它是内联的，则接口标头必须包含实现标头，并且通过它，实现的知识将“泄漏”到调用站点。

The implementation header foo_impl.h:

实现头文件 foo_impl.h：

#pragma once
#include "foo.h"

class Simple_foo : public Foo
{
    void do_it() override;
};

class Fancy_foo : public Foo
{
    void do_it() override;
};

Nothing special, just override the virtual functions of the interface. Because this exaple is simple, I put both implementations in the same files. In real applications that will be different.

没什么特别的，只是覆盖了接口的虚函数。因为这个例子很简单，我把两个实现放在同一个文件中。在实际应用中，情况会有所不同。

The implementation foo_impl.cpp:

foo_impl.cpp 的实现：

#include "foo_impl.h"
#include <iostream>

void Simple_foo::do_it()
{
    std::cout << "simple foo\n";
}

void Fancy_foo::do_it()
{
    std::cout << "fancy foo\n";
}

Just implement the functions.

只需实现功能。

The main.cpp:

main.cpp：

#include "foo.h"

int main()
{
    auto sf = create_foo(Implementations::Simple);
    sf->do_it();
    auto ff = create_foo(Implementations::Fancy);
    ff->do_it();
    return 0;
}

Through the enum we can select the implementation we want. The pointers are of type Foo::Ptr, an alias for std::unique_ptr<Foo>. The callsite has no knowledge of the implementation at all, only the interface.

通过枚举我们可以选择我们想要的实现。指针的类型是 Foo::Ptr，是 的别名std::unique_ptr<Foo>。调用站点根本不知道实现，只知道接口。

The output will be as expected:

输出将如预期的那样：

simple foo
fancy foo