如何用C创建的指针数组++ [英] How to create array of pointers in C++

问题描述

在类节点：

 类节点{
上市：
    int数据;
    INT numchild;
    节点**节点列表;    节点（int数据，int类型）;
};
 

我要指针（节点列表）到其他节点，其中有从该节点的边缘。

的数组

是构建这样一个数组（并宣布它上面的方法），正确的和最佳（或最简单的）可能的方式的下列方式？如果不是，为什么和最新最好的方法是什么？

 节点::节点（INT D，int类型）{
    数据= D;
    numchild =秒;
    节点列表=新节点* [S]。
}
 

解决方案

通常你不应该推倒重来。原阵列几乎总是错的路要走。看一看从STL的各种容器，即的std ::矢量，的std ::列表。我woud想在你的情况下，的std ::矢量可能是最好的解决方案。

如果你想坚持与原阵，一个快速的警告：新节点* [S] 不初始化数组，这样的内容将是不确定的。但是，如果你添加一组括号（新节点* [S]（））将被初始化为零这是一件好事，因为它可以帮助你发现哪些条目有已经被填满。

另外，你目前的code缺乏析构函数再次删除[] 的节点列表。这正是为什么推荐使用标准容器：他们有析构函数，将做的工作适合你。

In class Node:

class Node {
public:
    int data;
    int numchild;  
    Node** nodelist;

    Node(int data, int s);
};

I want an array of pointers (nodelist) to other nodes, which have edges from this node.

Is the following way of constructing such an array (and the above way of declaring it) a correct and the best (or easiest) way possible? If not, why and whats the best way?

Node::Node(int d, int s) {
    data = d;
    numchild = s;
    nodelist = new Node*[s];
}

解决方案

Generally you shouldn't reinvent the wheel. Raw arrays are almost always the wrong way to go. Have a look at the various containers from the STL, i.e. std::vector, std::list. I woud guess that in your case a std::vector might be the best solution.

If you want to stick with the raw array, a quick warning: new Node*[s] does not initialize the array, so the contents will be undefined. But if you add a set of parentheses (new Node*[s]()) it will be initialized to zero which is a good thing as it helps you spot which entries have been filled already.

Also, your current code lacks a destructor to delete[] the nodelist again. That's exactly why the use of standard containers is recommended: they have destructors that will do the work for you.

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