C ++递归类型定义是可能的，特别是可以将向量< T>在T的定义内？

问题描述

对于我的一个项目，我真正想要做的是（将其简化到最低限度）;

For one of my projects, what I really wanted to do was this (simplifying it to the bare minimum);

struct Move
{
    int src;
    int dst;
};

struct MoveTree
{
    Move move;
    std::vector<MoveTree> variation;
};

我必须承认，我认为不可能直接这样做，在MoveTree中的MoveTree的向量将是verboten。但我试过它反正，它的作品精美。我使用的是Microsoft Visual Studio 2010 Express。

I must admit that I assumed that it wouldn't be possible to do this directly, I thought a vector of MoveTree s within a MoveTree would be verboten. But I tried it anyway, and it works beautifully. I am using Microsoft Visual Studio 2010 Express.

这是可移植的吗？这是好的做法吗？我有什么要担心的吗？

Is this portable ? Is it good practise ? Do I have anything to worry about ?

编辑：我问了第二个

I've asked a second question hoping to find a good way of doing this.

推荐答案

C ++标准（2003）清楚地说，实例化一个具有不完全类型的标准容器，调用未定义的行为。

The C++ Standard (2003) clearly says that instantiating a standard container with an incomplete type, invokes undefined-behavior.

在§17.4.3.6/ 2中，

The spec says in §17.4.3.6/2,

__ [..]

- 如果在实例化模板组件时将不完全类型（3.9）用作模板参数。

__ [.. ]

__ [..]
— if an incomplete type (3.9) is used as a template argument when instantiating a template component.
__ [..]

这篇关于C ++递归类型定义是可能的，特别是可以将向量< T>在T的定义内？的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持！