Question

在以下代码中：

class A
{
    void aMethod() { }
    void aConstMethod() const { }
}

class B
{
    const A a; // Not initialized in the constructor, but at a latter time

    void initA()
    {
        a = new A(); // Error: can only initialize const member 'a' inside constructor
    }

    void doStuff()
    {
        //a.aMethod(); shouldn't be allowed to call this here, B can only read from A.
        a.aConstMethod();
    }
}

我希望课程B只能从const调用immutable或A方法。但是，B只能在构造后创建A的实例，因此无法在构造函数中初始化A。我是否可以修改上述代码，而无需从var const移除a？

Answer 1

使用std.typecons.Rebindable：

class A
{
    void aMethod() { }
    void aConstMethod() const { }
}

class B
{
    import std.typecons: Rebindable;

    Rebindable!(const A) a; // Not initialized in the constructor, but at a latter time

    void initA()
    {
        a = new A(); // Error: can only initialize const member 'a' inside constructor
    }

    void doStuff()
    {
        static assert(!__traits(compiles, a.aMethod())); // shouldn't be allowed to call this here, B can only read from A.
        a.aConstMethod();
    }
}

Init const对象在构造函数之外

1 个答案: