如何使用Boost C到遍历的递归变型矢量++ [英] How to traverse a recursive variant vector using Boost C++

问题描述

我想建立一个矢量树结构如下：

I want to build a vector tree constructed as follows

struct myStruct {
    int a;
    string b;
};

typedef boost::make_recursive_variant<
      myStruct *
    , std::vector< boost::recursive_variant_ >
    >::type myStruct_tree;

如果我追加多个向量，向量和这些载体，将如何使用某种位置矢量的我遍历树

If I append a multiple vectors, and vectors in those vectors, how would I traverse the tree using some kind of position vector

vector<int>

将查找在矢量指针树中定义每个矢量/子向量的对象的位置。

Which locates the objects position in each vector/subvector defined in the vector pointer tree.

推荐答案

我不是100％肯定我明白你的问题，所以如果我得到的东西错了，请澄清。我假定的载体是路径，每个元件使移动到元素的索引。考虑到这一点，我想尝试这样的：

I'm not 100% sure I understand your question, so please clarify if I get something wrong. I'm assuming the vector is a path, with each element giving the index of the element to move to. With that in mind, I'd try something like this:

myStruct_tree t = ...;

myStruct_tree* tit = &t;
for(iterator pit=path.begin(); pit!=path.end(); ++pit)
{
    // retrieve branches at current position
    vector<myStruct_tree>& vec = get<2>(*tit);

    // get next position in tree, will throw on bad index
    tit = &vec.at(*pit);
}
assert(tit);

// retrieve the leaf at the final position
myStruct* res = get<1>(*tit);

注：

• 使用一个无符号整数的指数会多一点自然的我。


• 矢量::在（）抛出对超出范围的异常。


• 我没有抬头的如何获得变种的元素确切的语法。随着版本获取。1 GT;（*针锋相对）; 我上面使用，我的意思是检索的第一个元素（即MYSTRUCT指针）之一。另外，它应该是一个返回参考或以处​​理错误引发之一。


• 考虑更换，以提供更好的诊断用自己的code中的检查范围，。我没有做到这一点，保持清晰的算法的结构。

• Using an unsigned integer for the indices would be a bit more natural to me.
• vector::at() throws an exception on out-of-range.
• I haven't looked up the exact syntax of how to get an element of the variant. With the version get<1>(*tit); I used above, I meant the one that retrieves the first element (i.e. the myStruct pointer). Further, it should be the one that returns a reference or throws in order to handle errors.
• Consider replacing the range-checking with your own code in order to provide better diagnostics. I didn't do it to keep the structure of the algorithm clear.

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