插入新节点 [英] Inserting New Node

问题描述

这是我正在写的一个程序的功能，可以更熟悉节点.

Here is a function of a program I'm writing to get more familiar with nodes.

它将创建一个新节点，并将信息插入到代码字段中，然后指向现有的第一个节点.然后将头分配给新创建的节点；

It creates a new node and inserts the information into the code field and then points to the existing first node. Then assigns the head to the newly created node;

不幸的是，它给了我new_node-> location = code不兼容的类型；

Unfortunately it's giving me a incompatible types for new_node->location = code;

typedef char LibraryCode[4];

typedef struct node {
    LibraryCode location;
    struct node *next;
} Node;

void insertFirstNode(LibraryCode code, Node **listPtr) {
    Node *new_node;
    new_node=malloc(sizeof(Node));
    new_node->location = code;
    new_node->next = *listPtr;
    *listPtr = new_node;
}

推荐答案

LibraryCode 被 typdef 用作 char [4] .您不能只将一个数组分配给另一个数组，而是需要将 memcpy 或 strcpy 数据从一个数组分配到另一个数组.

LibraryCode is typdefed as a char [4]. You can't just assign one array to another, you'll need to memcpy or strcpy the data from one array to the other.

一个简单的例子:

void foo() {
   char a[4];
   char b[4];
   a = b;
}

编译此错误:

In function ‘foo’:
error: incompatible types when assigning to type ‘char[4]’ from type ‘char *’
   a = b;
     ^

您可以看到您实际上是在尝试为数组分配指针，这是不兼容的类型.

You can see that you're actually trying to assign a pointer to the array, which are incompatible types.

typedef char LibraryCode [4]; 可能不是一个好主意.如果您要在节点中为字符串保留一个固定大小的缓冲区，那么我将放弃 typedef ，这样您就可以清楚地知道自己在做什么.另外，我绝不会将 char [4] 按值传递给函数.对于字符串，请传递 const char * .

The typedef char LibraryCode[4]; is probably not a good idea anyway. If you're going to keep a fixed-size buffer for the string in your node, then I would just ditch the typedef so it's clear what you're doing. Also, I would never pass a char [4] by value to a function. For strings, pass a const char*.

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