尝试用C语言创建链表

我正在尝试用 C 语言创建一个链表结构。但我不太确定出了什么问题。我的错误是：

linked.c:6:2: error: unknown type name ‘linkedList’
linked.c: In function ‘makeList’:
linked.c:30:2: error: ‘first’ undeclared (first use in this function)
linked.c:30:2: note: each undeclared identifier is reported only once for each function it appears in
linked.c: In function ‘addToList’:
linked.c:36:9: error: used struct type value where scalar is required
linked.c:43:13: error: incompatible types when assigning to type ‘int *’ from type ‘linkedList’

如果有人能看到问题所在并向我解释，我将不胜感激。我的代码如下。

#include <stdio.h>

typedef struct linkedList
{
        int first;
        linkedList* rest;
} linkedList;

linkedList makeList(int a, int b, int c);
void addToList(linkedList* ll, int a);

int main()
{
        linkedList ll = makeList(1,3,5);
        addToList(&ll, 7);
        addToList(&ll, 9);
        return 0;
}

linkedList makeList(int a, int b, int c)
{
        linkedList ll;
        ll.first = a;
        linkedList second;
        second.first = b;
        linkedList third;
        third.first = c;
        third.rest = NULL;
        second.rest = &c;
        first.rest = &b;
        return first;
}

void addToList(linkedList* ll, int a)
{
        while (*ll)
        {
                if (ll->rest == NULL)
                {
                    linkedList newL;
                    newL.first = a;
                    newL.rest = NULL;
                    ll->rest = newL;
                    break;
            } else
            {
                    continue;
            }
    }
}

C编译器没有完整的typedef of linkedList在您尝试在您的应用程序中使用它之前struct。您有几个选择：

typedef struct linkedList
{
    int first;
    struct linkedList* rest;
} linkedList;

Or:

typedef struct linkedList linkedList;  // C allows this forward declaration

struct linkedList
{
    int first;
    linkedList* rest;
};

这是你的起点。

其他问题包括但不限于：

• Your makeList函数引用变量first但它似乎没有在任何地方定义。
• ll->rest = newL;分配一个类型linkedList to a pointer to linkedList (linkedList *）您不能将值分配给指向值的指针。编译器错误信息linked.c:43:13:...指出这一点。它需要是ll->rest = &newL;... 然而...
• newL对于函数来说是本地的addToList，因此您不能将其地址分配给持久列表项，因为当代码离开该块时它将超出范围。
• In addToList您正在将指向整数的指针分配给一个保存指向的变量linkedList, e.g., second.rest = &c;.