您可以在函数参数内部解释此空指针吗？

由六道仙人发布于 2020-05-09 14:11:21

这是有问题的代码！我没有花太多时间在C ++上，并且正在实现二进制搜索树。

void BST::Preorder(void(*visit)(const Node *))

稍后，该函数的调用方式如下：

bst.Preorder(PrintNode)

使用PrintNode实现如下：

void PrintNode(const Node* n)
{
    cout << n->GetValue() << ",";
}

被要求实现此Preorder函数-但是我对这里发生的事情感到非常困惑。谁能指出我的正确方向。谢谢。

2020-05-09 14:11:21

void BST::Preorder(void(*visit)(const Node *))

This is Preorder function of BST class.
The parameter of that function is a function pointer.
*visit means it's a function pointer that accept a const Node * as its own paremeter and return void.

So generally you would have to implement that visit function elsewhere. Then uses that visit function somewhere in Preorder function.

2020-05-09 14:11:21

This is a parameter of type pointer to function accepting const Node * and returns void and the name of the parameter is visit.

In other words if you have a function like func(), you can pass it to BST::Preorder()

void func(const Node * )
{
    //do something
}

这家伙很懒，什么都没留下