取消引用C ++中的void指针 [英] Dereferencing the void pointer in C++

问题描述

我正在尝试实现通用链表.节点的结构如下-

I'm trying to implement a generic linked list. The struct for the node is as follows -

typedef struct node{
        void *data;
        node *next;      
};

现在，当我尝试为数据分配地址时，例如假设一个int，如-

Now, when I try to assign an address to the data, suppose for example for an int, like -

int n1=6;
node *temp;
temp = (node*)malloc(sizeof(node));
temp->data=&n1;

如何从节点获取n1的值?如果我说-

How can I get the value of n1 from the node? If I say -

cout<<(*(temp->data));

我明白了-

`void*' is not a pointer-to-object type 

当我为它分配一个int地址时，void指针不会被强制转换为int指针类型吗?

Doesn't void pointer get typecasted to int pointer type when I assign an address of int to it?

推荐答案

必须首先将void*转换为指针的实际有效类型(例如int*)，以告诉编译器您希望取消引用多少内存.

You must first typecast the void* to actual valid type of pointer (e.g int*) to tell the compiler how much memory you are expecting to dereference.

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