Question

我正在尝试从用户那里获取一个数字并创建2个链接列表，一个列表仅包含偶数，另一个列表包含奇数，我必须返回已插入的总数，并通过引用传递链接列表。我的主要问题是返回2个新的链表，这是我到目前为止所做的。我得到一个错误错误C2440“功能”：无法从“节点**”转换为“节点”。

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

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

    struct node* head;
    int get_values(node,node);

    void main() {
        node *even_node,*odd_node;
        get_values(&even_node,&odd_node);

    }
    int get_values(node *even, node *odd) {
        int value, counter_total = 0;
        node  *curr_even;
        node  *curr_odd;
        head = NULL;
        printf("enter value:");
        scanf_s("%d", &value);
        if (value == -1) {
            return NULL;
        }
        if (value % 2 == 0) {
            even = (node*)malloc(sizeof(node));
            curr_even = even;
            even->data = value;
            counter_total++;
        }
        else {
            odd = (node*)malloc(sizeof(node));
            curr_odd = odd;
            odd->data = value;
            counter_total++;
        }
        //2nd and on insertion.
        while (value != -1) {
            printf("enter a value positive value");
            scanf_s("%d", &value);
            if (value == -1) {
                curr_even->next = NULL;
                curr_odd->next = NULL;
                break;
            }

            else if (value % 2 == 0) {
                curr_even->next = (node *)malloc(sizeof(node));
                curr_even = curr_even->next;
                curr_even->data = value;//current value
                counter_total++;
            }
            else {
                curr_odd->next = (node*)malloc(sizeof(node));
                curr_odd = curr_odd->next;
                curr_odd->data = value; //current value
                counter_total++;
            }

            return counter_total;
        }
    }

Answer 1

您的代码有很多错误。

getvalues的函数定义应该有一个双指针。
在函数中，您malloc正在使用函数参数。当您需要分配局部变量并将其添加到列表中时。
一次do while循环就足够了，您将添加不必要的代码重复。

请参阅下面的固定代码

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

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

struct node* head;
int get_values(node **,node **);

void main() {
    node *even_node= NULL,*odd_node= NULL;
    get_values(&even_node,&odd_node);

}
int get_values(node **even, node **odd) {
    int value, counter_total = 0;
    node  *curr_even;
    node  *curr_odd;
node  *new_node;

do
{
  printf("enter value:");
  scanf("%d", &value);
  if (value == -1) {
      return counter_total;
  }
  if (value % 2 == 0) 
  {
      new_node = (node*)malloc(sizeof(node));
      new_node -> data = value;
      new_node -> next = NULL;
      if (*even == NULL)
      {
        *even = new_node;
        curr_even = *even;        
      }
      else
      {
         curr_even ->next = new_node;
         curr_even = curr_even -> next;
      }
      counter_total++;
  }
  else 
  {
      new_node = (node*)malloc(sizeof(node));
      new_node -> data = value;
      new_node -> next = NULL;

      if (*even == NULL)
      {
        *even = new_node;
        curr_even = *even;        
      }
      else
      {
        curr_even ->next = new_node;
        curr_even = curr_even -> next;
      }
      counter_total++;
  }
}while (1);
}

Answer 2

如果您想通过引用获得通过，可以按照以下步骤进行。

import React from 'react';
import { TouchableOpacity } from 'react-native';
import PropTypes from 'prop-types';
import _ from 'lodash';

const TouchableDebounce = props => (
  <TouchableOpacity
    {...props}
    onPress={
      _.debounce(props.onPress, props.debounceTime, props.options)
    }
  >
    {props.children}
  </TouchableOpacity>
);

TouchableDebounce.propTypes = {
  onPress: PropTypes.func.isRequired,
  children: PropTypes.element,
  options: PropTypes.objectOf(PropTypes.any),
  debounceTime: PropTypes.number,
};
TouchableDebounce.defaultProps = {
  children: null,
  options: { 'leading': true, 'trailing': false },
  debounceTime: 500,
};
export default TouchableDebounce;

使用链表功能将偶数和奇数分开

2 个答案: