C++ 将数组返回到链表

发布于08月26日

// Implements a list of numbers using a linked list

#include <cs50.h>
#include <stdio.h>
#include <stdlib.h>
#include <string.h>

typedef struct node
{
    int number;
    struct node *next;
}
node;

node *makelist(int argc, char *argv[]);
int check_spaces(char *input);
int *nodes(char *input, int amount);
node *add(char *input, node *list, int *numbers, int amount);


/*int delete();
int find();*/

int main(int argc, char *argv[])
{
    //Get linked list
    node *list = makelist(argc, argv);
    //Print linked list
    for (node *ptr = list; ptr != NULL; ptr = ptr->next)
    {
        printf("%i ", ptr->number);
    }
    printf("\n");
    // Free memory
    node *ptr = list;
    while (ptr != NULL)
    {
        node *next = ptr->next;
        free(ptr);
        ptr = next;
    }
    char c;
    //Give choices
    printf("Do you want to (d)delete a node, (a)add a node, or (f)find a node? ");
    //Repeat if incorrect input
    do
    {
        scanf("%c", &c);
        if (c == 'd' || c == 'a' || c == 'f')
        {
            break;
        }
        else
        {
            printf("Wrong input, try again. ");

        }
    }
    while (c != 'd' || c != 'a' || c != 'f');

    //If user chose to "add" node(s)
    if (c == 'a')
    {
        char *name = get_string("");
        char *input = get_string("What are the node(s) that you want to add? ");
        int amount = check_spaces(input);
        int *numbers = malloc(amount * 4);
        *numbers = *nodes(input, amount);
        list = add(input, list, numbers, amount);
        for (node *ptr2 = list; ptr2 != NULL; ptr2 = ptr2->next)
        {
            printf("%i ", ptr2->number);
        }
        free(numbers);
        free(list);
    }
    //If user chose to "delete" node(s)
    else if (c == 'd')
    {
        printf("d");
    }
    //If user chose to "find" node(s)
    else
    {
        printf("c");
    }
    return 0;
}
//Make linked list
node *makelist(int argc, char *argv[])
{
    // Memory for numbers
    node *list = NULL;

    // For each command-line argument
    for (int i = 1; i < argc; i++)
    {
        // Convert argument to int
        int number = atoi(argv[i]);

        // Allocate node for number
        node *n = malloc(sizeof(node));
        if (n == NULL)
        {
            return 0;
        }
        n->number = number;
        n->next = NULL;

        // If list is empty
        if (list == NULL)
        {
            // This node is the whole list
            list = n;
        }

        // If list has numbers already
        else
        {
            // Iterate over nodes in list
            for (node *ptr = list; ptr != NULL; ptr = ptr->next)
            {
                // If at end of list
                if (ptr->next == NULL)
                {
                    // Append node
                    ptr->next = n;
                    break;
                }
            }
        }
    }
    return list;
}
//Add function
node *add(char *input, node *list, int *numbers, int amount)
{
    int a = 0;
    while (numbers[a] != '\0')
    {
        a++;
        node *n = malloc(sizeof(node));
        if (n == NULL)
        {
            return 0;
        }
        n->number = numbers[amount-a];
        // Prepend node to list
        n->next = list;
        list = n;
    }
    return list;
}
//Get command line arguments from printf
int *nodes(char *input, int amount)
{
    int *y = malloc(amount * 4);
    char *x = &input[0];
    y[0] = atoi(x);
    int i = 0;
    int j = 1;
    while (input[i] != '\0')
    {
        if (input[i] == ' ')
        {
            x = &input[i+1];
            y[j] = atoi(x);
            j++;
        }
        i++;
    }
    return y;
}
//How many nodes are we messing with
int check_spaces(char *input)
{
    int numbers = 1;
    int i = 0;
    while (input[i] != '\0')
    {
        if (input[i] == ' ')
        {
            numbers++;
        }
        i++;
    }
    return numbers;
}

/*int delete()
{

}
int find()
{

}*/

该问题源于C无法返回array.问题是关于添加函数，因为其他函数是不完整的.

我的整个计划是使用atoi将用户的命令行参数转换为数字，然后反向将它们添加到链表中.如果我的 list 是3，4，5，我加上1，2，它就会变成1，2，3，4，5(理论上).

我的 node 函数执行所有的转换，由于C的规则，我不能返回数组，所以我返回了一个指针.我认为所有的数字都是背靠背的，因为有了Malloc，所以我有一个循环遍历n-&gt；Numbers=Numbers[Amount-a].第一个值是正确的，但然后其他的值只是一堆垃圾值.

我很确定这是因为指针只指向第一个对象，其余的对象都是按添加的顺序排列的.我曾try 过访问其他值，但没有这样的运气.老实说，我觉得我错过了一些重要的东西.

一定有什么事情我在这里过分复杂化了.

推荐答案

restarting things: a C complete EXAMPLE

关于行参数

链表

重写:一个 node

和一份名单

一些规划

一些函数

更多功能

第二个项目

create and destroy list

A third program

inserting data: 添加()

To test this we need display()

第4个节目现在甚至可以显示一些数据

删除() and 发现() simple implementations

第五期节目时间到了

插入命令行参数(如果有)

这句话是唯一的 fresh 事:

输出

到目前为止的完整例子

C code

C++相关问答推荐

restarting things: a `C` complete EXAMPLE

`create` and `destroy` list

A third `program`

inserting data: `添加()`

To test this we need `display()`

`删除()` and `发现()` simple implementations

`C` code