获取常量引用的迭代器 [英] Get iterator for const reference

问题描述

我正在开发一个必须使用 begin() 方法返回迭代器的类.此外，我必须开发一个函数来接收此类的 const 引用并对其进行迭代.

I'm developing a class that must return an iterator with the begin() method. Also, I have to develop a function that receives a const reference of this class and iterates over it.

当我尝试从该方法获取迭代器时，出现以下编译错误:对象具有与成员函数不兼容的类型限定符."我不明白为什么出现此错误.

When I try to get an iterator from this method, I have the following compilation error: "the object has type qualifiers that are not compatible with the member function."I can't understand why this error appears.

这是我写的代码:

// ------------ class Neuron -------------
class Neuron { ... };
// ---------------------------------


// ------------ class AbstractLayer -------------
class AbstractLayer {
public:

    class Iterator : public std::iterator<std::input_iterator_tag, Neuron> {
        public:
            Iterator(Neuron *neurons) : _neurons(neurons) {}
        private:
            Neuron *_neurons;
    };

    virtual Iterator begin() = 0;
    virtual const Iterator begin2() = 0;
};
// ----------------------------------------


// ------------ class Layer -------------
class Layer : AbstractLayer {

public:
    Layer(){};
    Iterator begin(){ return Iterator(_neurons); }
    const Iterator begin2(){ return (const Iterator)begin(); }

private:
    Neuron *_neurons;
    int _size;
};
// --------------------------------


// ------------ Method where the problem is -------------------
void method(const AbstractLayer &layer){
    // Error in both methods: 
    // "the object has type qualifiers that are not compatible with the member function."
    layer.begin();
    layer.begin2();
}
// -------------------------------------------------------------

推荐答案

在 method 函数中，layer 引用了一个 constant 对象.这意味着您只能调用标记为 const 的函数.像例如

In the method function, layer references a constant object. That means you can only call functions marked as const. Like e.g.

class AbstractLayer {
public:
    ...
    virtual const Iterator begin() const = 0;  // <- Note use of `const` here
    ...
};

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