重载 C++ 时运行特定代码运营商

发布于 2024-10-08 08:40:59 字数 1176 浏览 0 评论 0原文

我有一个类，我们称之为 Foo，它包含以下 3 个方法（重载 左关联 < 二元运算符）：

... operator<(A a) { return *this; }
... operator<(B b) { return *this; }
... operator<(C c) { return *this; }

A、B、C 类不是以任何方式相关（如果是的话，有什么关系）。

现在在我的程序中我只有以下两种情况：

A a = new A();
B b = new B();
C c = new C();

(First Case): new Foo() < a < b;

或者

(Second Case): new Foo() < a < b < c;

每当我有第一种情况（以 b 结尾）时，我想在阅读后执行函数 run() （我知道） b 实例。所以我的想法是，我将在 Foo 类中添加以下代码：

... operator<(B b)
{
    run();
}

现在，当我获得第一种情况的代码时，将执行 run() 。

问题是，当我有类似于第二种情况的代码时（以 c 结尾）。我想再次执行 run() 函数，但直到我知道 c 是什么。因此，如果我有前面的代码 run() 将在执行 时被调用b 这不是我想要的，因为我还不知道c。如果我在 operator<(C c) 中添加 run()，我将调用 run() 两次。

简而言之，我想要完成的是当第一种情况在 operator<(B b) 处调用 run() 时，当我有第二种情况时仅调用 run在操作符<(C c)处。

关于如何解决这个问题有什么想法（如果可以的话）？

原文

I have a class, let's call it Foo, which contains the 3 following methods (overloading the left-associative < binary operator):

... operator<(A a) { return *this; }
... operator<(B b) { return *this; }
... operator<(C c) { return *this; }

A, B, C classes are not related in any way(if that, is of any matter).

Now In my program I only have the 2 following cases:

A a = new A();
B b = new B();
C c = new C();

(First Case): new Foo() < a < b;

(Second Case): new Foo() < a < b < c;

Whenever I have the first case (which ends with b), I want to execute a function run() when I have read(I know) the b instance. So the idea is that I would have the following code in the Foo class:

... operator<(B b)
{
    run();
}

Now when I have the code of the first case, run() is being executed.

The problem is that when I have code like in the second case(which ends in c).
I want to execute the run() function again but NOT until I know what c is.
So If I have the previous piece of code run() will be called when doing < b
which is not what I want as I don't know c yet.
If I add run() in operator<(C c) I will call run() twice.

In a few words what I want to accomplish is when having the first case call run() at operator<(B b) and when I have the second case ONLY call run at operator<(C c).

Any ideas on how this can be solved(if it can)?

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橘香 2024-10-15 08:40:59

您可以创建单独的代理对象返回类型，因此当您为该代理对象定义运算符时，您知道在它之前有哪些类型。

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满身野味 2024-10-15 08:40:59

class PreFoo {
public:
    PreFoo() { ... }
    PreFoo & operator<<(A a) { ...; return *this; }
    PreFoo & operator<<(B b) { ...; return *this; }
    PreFoo & operator<<(C c) { ...; return *this; }

    int aa, bb, cc; // save your information
};

PreFoo MakeFoo() { return PreFoo(); }

class Foo {
public:
    Foo(const PreFoo & pre) { run(); } // implicit conversion

    void run() {} // your run!!
};


void g() {
  A a; B b; C c;
  // implicit conversion at the end
  Foo foo1 = MakeFoo() << a << b;
  Foo foo2 = PreFoo() << a << b << c;
}

class PreFoo {
public:
    PreFoo() { ... }
    PreFoo & operator<<(A a) { ...; return *this; }
    PreFoo & operator<<(B b) { ...; return *this; }
    PreFoo & operator<<(C c) { ...; return *this; }

    int aa, bb, cc; // save your information
};

PreFoo MakeFoo() { return PreFoo(); }

class Foo {
public:
    Foo(const PreFoo & pre) { run(); } // implicit conversion

    void run() {} // your run!!
};


void g() {
  A a; B b; C c;
  // implicit conversion at the end
  Foo foo1 = MakeFoo() << a << b;
  Foo foo2 = PreFoo() << a << b << c;
}

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