如何用OpenAI ChatGPT进行编码
这篇文章主要介绍“如何用OpenAI ChatGPT进行编码”,在日常操作中,相信很多人在如何用OpenAI ChatGPT进行编码问题上存在疑惑,小编查阅了各式资料,整理出简单好用的操作方法,希望对大家解答”如何用OpenAI ChatGPT进行编码”的疑惑有所帮助!接下来,请跟着小编一起来学习吧!
ChatGPT 是一种以类似聊天的方式进行交互的人工智能网络,从名字上看起来像是另一个“聊天机器人”,但实际上已经远超聊天的范畴了,当要求它编写一些代码时,结果相等震撼,它可以写出一些意想不到的连贯代码,而这些代码对于开发者很有启发。
开始要求它使用 JavaScript 编写一个散列(HASH)密码的函数。具体一点就是编写一个 JavaScript 函数,它接受密码的字符串输入并输出散列(HASH)密码。
function hashPassword(password) { return crypto.createHash("sha256").update(password).digest("hex");}
然后逐步提高编码的难度,实现一些更加符合项目开发的需求,于是输入:
编写一个在请求正文中包含用户名和密码的 Express 路由处理程序,对密码进行哈希处理,在 MySQL 数据库中查找现有用户并检查哈希后的密码是否匹配。
const express = require("express");const bcrypt = require("bcrypt");const mysql = require("mysql");// 创建 MySQL 数据库连接const db = mysql.createConnection({ host: "localhost", user: "root", password: "", database: "mydb",});// 连接数据库db.connect((err) => { if (err) { throw err; } console.log("MySQL connected...");});// 创建 express appconst app = express();// 路由 handlerapp.post("/login", (req, res) => { // get username and password from request body const { username, password } = req.body; // HASH 密码 const hashedPassword = bcrypt.hashSync(password, 10); // 在数据库中查找用户名称 let sql = `SELECT * FROM users WHERE username = '${username}'`; let query = db.query(sql, (err, result) => { if (err) throw err; // 检查密码是否匹配 if (bcrypt.compareSync(result[0].password, hashedPassword)) { res.send("Login successful!"); } else { res.send("Incorrect username or password"); } });});
这段 JavaScript 代码基本是没什么问题的,可以直接复制到项目中进行调试。接下来使用别的语言来试一下:
用C语言编写一个链表结构的代码,具有根据“标识”属性在链表中添加、删除和查找节点的功能。
#include <stdio.h>#include <stdlib.h>// Structure for a node in the linked listtypedef struct Node { int identity; // Property of the node struct Node *next; // Pointer to the next node} Node;// Function to add a node to the linked listvoid addNode(Node **head, int identity) { // Create a new node Node *newNode = (Node*)malloc(sizeof(Node)); // Set the identity of the new node newNode->identity = identity; // Set the next pointer of the new node to point to the head newNode->next = *head; // Set the head to point to the new node *head = newNode;}// Function to delete a node from the linked listvoid deleteNode(Node **head, int identity) { // Create a temporary node to traverse the linked list Node *temp = *head; // Check if the head is the node to be deleted if (temp != NULL && temp->identity == identity) { // Set the head to point to the next node *head = temp->next; // Free the memory allocated for the node free(temp); return; } // Traverse the linked list to find the node to be deleted while (temp != NULL && temp->identity != identity) { // Set the previous node to the current node Node *prev = temp; // Set the current node to the next node temp = temp->next; } // Check if the node was found if (temp == NULL) return; // Set the previous node to point to the next node prev->next = temp->next; // Free the memory allocated for the node free(temp);}// Function to find a node in the linked listNode* findNode(Node *head, int identity) { // Create a temporary node to traverse the linked list Node *temp = head; // Traverse the linked list to find the node while (temp != NULL) { // Check if the current node has the given identity if (temp->identity == identity) { return temp; } // Set the current node to the next node temp = temp->next; } // Return NULL if the node was not found return NULL;}int main() { // Create a head node for the linked list Node *head = NULL; // Add some nodes to the linked list addNode(&head, 1); addNode(&head, 2); addNode(&head, 3); // Find a node in the linked list Node *node = findNode(head, 2); // Print the identity of the node printf("Node with identity 2: %d\n", node->identity); // Delete a node from the linked list deleteNode(&head, 2); // Find the deleted node in the linked list node = findNode(head, 2); // Check if the node was deleted if (node == NULL) { printf("Node with identity 2 was deleted\n"); } return 0;}
到此,关于“如何用OpenAI ChatGPT进行编码”的学习就结束了,希望能够解决大家的疑惑。理论与实践的搭配能更好的帮助大家学习,快去试试吧!若想继续学习更多相关知识,请继续关注编程网网站,小编会继续努力为大家带来更多实用的文章!
免责声明:
① 本站未注明“稿件来源”的信息均来自网络整理。其文字、图片和音视频稿件的所属权归原作者所有。本站收集整理出于非商业性的教育和科研之目的,并不意味着本站赞同其观点或证实其内容的真实性。仅作为临时的测试数据,供内部测试之用。本站并未授权任何人以任何方式主动获取本站任何信息。
② 本站未注明“稿件来源”的临时测试数据将在测试完成后最终做删除处理。有问题或投稿请发送至: 邮箱/279061341@qq.com QQ/279061341