使用boost ::可选恒类型 - C ++ [英] Using boost::optional with constant types - C++

问题描述

我有一个使用的boost ::可选的来保存值的容器类。这里是code样子，

I have a container class which uses boost::optional to hold the value. Here is the code looks like,

template<typename T>
struct traits
{
    typedef T  value_type;
    typedef T& reference;
};

template<typename T>
struct traits<const T>
{
    typedef const T  value_type;
    typedef const T& reference;
};

template<typename T>
struct traits<T*>
{
    typedef T* value_type;
    typedef T* reference;
};

template<typename T>
struct traits<const T*>
{
    typedef const T* value_type;
    typedef const T* reference;
};

template<typename T>
class container
{
public:

    typedef typename traits<T>::reference reference;
    typedef typename traits<T>::value_type value_type;

    container() {}

    void set(reference value) {
        op.reset(value);
    }

    reference get() const {
        return boost::get(op);
    }

private:
    boost::optional<value_type> op;
};

int main()
{
    foo f;
    container<const foo> c;
    c.set(f);
    return 0;
}

它可以很好地用于其它类型除了常量。当我使用常量类型我收到错误（常量富* 正常工作）。

It works well for other types except const. I am getting error when I use const types (const foo* works fine).

1. 是的boost ::可选的支持定类型？如果没有，我怎么能解决这个问题？


2. 有一个现成的性状可实现，我可以使用，而不是定义我自己的特质？

1. Is boost::optional supports constant types? If no, how can I work around this issue?
2. Is there a ready made traits implementation available which I can use rather than defining my own traits?

任何帮助将是巨大的！

推荐答案

这个问题是不是与的boost ::可选的，但与逻辑你想要什么去做。首先，创建常量的一个容器，然后您尝试修改什么遏制。我会感到惊讶，如果是工作。

The problem is not with boost::optional, but with the logic of what you're trying to do. First you create a container of const, and then you try to modify what's contained. I would be surprised if that worked.

我想你也许应该做些什么标准集装箱（如矢量）做的，禁止不可复制的模板参数。

I think you should probably do what standard containers (like vector) do and forbid non-copyable template arguments.

否则，你就不得不忍受的事实，当 T 是设置方法不起作用不可复制，并提供执行初始化的构造：

Otherwise you'll have to live with the fact that your set method won't work when T is non-copyable, and provide a constructor that performs the initialization:

class container
{
public:

    container(reference init_value) : op(init_value) {}

};

int main()
{
    foo f;
    container<const foo> c(f);  // OK
    //   c.set(f);  NO
    return 0;
}

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