Cursor编程初体验，搭载GPT-4大模型，你的AI助手，自然语言编程来了

背景

这两天体验了下最新生产力工具Cursor，基于最新的 GPT-4 大模型，目前免费，国内可访问，不限次数，跨平台，你确定不来体验一把？官方的 Slogan ：

Build Software. Fast. Write, edit, and chat about your code with GPT-4
In partnership with OpenAI

在进行初步体验时，以下几点需要了解：

Cursor编程的上下文是你所在的代码文件，即如果你的文件后缀名为.py，那么在具体沟通时，就无需再出现“使用Python实现……”这种多余的前提，可以用更具体的限制；
代码生成之后还会提示合并还是丢弃：Accept或者Reject；
支持回滚：如果想撤销某次的Prompt，Ctrl+Z, Reject；
偶尔会出现生产的代码不完整的问题，只需问一句：没写完吧？
Ctrl+K，生成代码或者对选定的代码段进行优化；
Ctrl+L，对聊模式，对编程问题或者选定的代码段直接提问；

支持的语言

聊天模式

提示合并还是丢弃

以下通过12个简单的问题，从语言支持、语法支持、业务场景、代码解释、代码优化等方面来体验一把这个编程生产力利器。

理论上，对于复杂的任务，只要分解到GPT能够理解的那一步，都可以胜任。程序员的AI助手（它竟然可以帮我们找Bug~~），自然语言编程真的来了。至于下面生成的代码能不能用，还需要进行实际验证，但目测从语言支持、语义理解以及逻辑推理等方面来看，可用率至少80%吧。

语言支持

编程语言支持： TypeScript , HTML , CSS , Python , C , Rust , Go , CSharp , Java , PHP , SQL 等数十种。

Node.js

第一问（源自于我的一篇文章：使用bcryptjs对密码加密时，其校验原理是怎样的？）：
- 使用bcryptjs生成hash
- 校验密码

// Welcome to Cursor// 1. Try generating with command K on a new line. Ask for a new react component of an error popup.
// 2. Then, select the outputted code and hit chat. Ask if there's a bug. Ask how to improve.
// 3. Try selecting some code and hitting edit. Ask the bot to add a button that updates a statefield.
// 4. To try out cursor on your own projects, go to file menu (top left) and open a folder.
const bcrypt = require('bcryptjs');const password = 'myPassword123';
const salt = bcrypt.genSaltSync(10);
const hash = bcrypt.hashSync(password, salt);console.log(hash);const isMatch = bcrypt.compareSync('myPassword123', hash);
console.log(isMatch); // true

Python

第二问：
- 编写一个聚类算法，并提供实例

# Welcome to Cursor# 1. Try generating with command K on a new line. Ask for a pytorch script of a feedforward neural network
# 2. Then, select the outputted code and hit chat. Ask if there's a bug. Ask how to improve.
# 3. Try selecting some code and hitting edit. Ask the bot to add residual layers.
# 4. To try out cursor on your own projects, go to the file menu (top left) and open a folder.
# K-Means Clustering Algorithmimport numpy as np
import matplotlib.pyplot as pltclass KMeans:def __init__(self, k=2, max_iter=100):self.k = kself.max_iter = max_iterdef fit(self, X):self.centroids = {}# Initialize centroids randomlyfor i in range(self.k):self.centroids[i] = X[np.random.choice(len(X))]# Iterate until convergence or max iterations reachedfor i in range(self.max_iter):self.clusters = {}for j in range(self.k):self.clusters[j] = []# Assign each data point to the closest centroidfor x in X:distances = [np.linalg.norm(x - self.centroids[c]) for c in self.centroids]closest_centroid = distances.index(min(distances))self.clusters[closest_centroid].append(x)# Update centroids to be the mean of their assigned data pointsfor c in self.clusters:self.centroids[c] = np.mean(self.clusters[c], axis=0)def predict(self, X):predictions = []for x in X:distances = [np.linalg.norm(x - self.centroids[c]) for c in self.centroids]closest_centroid = distances.index(min(distances))predictions.append(closest_centroid)return predictions# Example usage
X = np.array([[1, 2], [1.5, 1.8], [5, 8], [8, 8], [1, 0.6], [9, 11]])
kmeans = KMeans(k=2)
kmeans.fit(X)
predictions = kmeans.predict(X)
colors = ['r', 'g', 'b', 'c', 'm', 'y']
for i in range(len(X)):plt.scatter(X[i][0], X[i][1], color=colors[predictions[i]])
plt.show()

Java

第三问（源自于一道面试题或者日常工作中遇到的一个小需求）：
- 请简单封装一个函数，实现友好时间的功能，要求该函数可输入一个参数【时间戳】，输出一个字符串【三天前、昨天、今天、明天、刚刚（10分钟内）】。

public static String friendlyTime(long timestamp) {long current = System.currentTimeMillis();long diff = current - timestamp;if (diff < 10 * 60 * 1000) {return "刚刚";} else if (diff < 24 * 60 * 60 * 1000) {return "今天";} else if (diff < 2 * 24 * 60 * 60 * 1000) {return "昨天";} else if (diff < 3 * 24 * 60 * 60 * 1000) {return "前天";} else {return "更早之前";}
}

Java

第四问（源自于一道面试题或者日常工作中遇到的一个小需求）：
- 请将一维数组的内容进行正向排序，并进行反向输出。[1，7，8，4，3，6，2，5，9]

public class Array {public static void main(String[] args) {int[] arr = {1, 7, 8, 4, 3, 6, 2, 5, 9};Arrays.sort(arr); // sort the array in ascending orderfor (int i = 0; i < arr.length; i++) {System.out.print(arr[i] + " ");}System.out.println();for (int i = arr.length - 1; i >= 0; i--) {System.out.print(arr[i] + " ");}}
}

Java

第五问（源自于一道面试题或者日常工作中遇到的一个小需求）：
- 请将两个数组的交集与并集取出，存放于新的数组中。[q,w,e,r,t],[a,w,d,s,r,g]

public static void main(String[] args) {String[] arr1 = {"q", "w", "e", "r", "t"};String[] arr2 = {"a", "w", "d", "s", "r", "g"};Set<String> set1 = new HashSet<>(Arrays.asList(arr1));Set<String> set2 = new HashSet<>(Arrays.asList(arr2));// IntersectionSet<String> intersection = new HashSet<>(set1);intersection.retainAll(set2);System.out.println("Intersection: " + intersection);// UnionSet<String> union = new HashSet<>(set1);union.addAll(set2);System.out.println("Union: " + union);
}

SQL

数据表结构

第六问：
- 连接本地数据库：localhost:3306，用户名：root，密码：root，从数据库poetry中的poems读取前10条数据并打引，通过author_id关联poems_author表。

USE poetry;
SELECT poems.title, poems.content, poems_author.name 
FROM poems 
JOIN poems_author 
ON poems.author_id = poems_author.id 
LIMIT 10;

语法支持

第七问：
- 编码实现从数组中找出最小值：[8, 1, 4, 0, -3]
- 采用lambda方式实现

public class Min {public static void main(String[] args) {int[] arr = {8, 1, 4, 0, -3};int min = Arrays.stream(arr).min().getAsInt();System.out.println("The minimum value in the array is: " + min);}
}

业务场景

第八问（源自于实际的物联网开发场景）：
- 使用YoLoV5编码实现识别烟雾火焰

import torch
from PIL import Image
import requests
from io import BytesIO# Load model
model = torch.hub.load('ultralytics/yolov5', 'yolov5s', pretrained=True)# Define class names
class_names = ['smoke', 'fire']# Define function to predict image
def predict_image(image_url):# Load image from URLresponse = requests.get(image_url)img = Image.open(BytesIO(response.content))# Predict imageresults = model(img)# Get class predictionsclass_predictions = results.pred[0].softmax(-1)# Get class indicesclass_indices = class_predictions.argmax(-1).tolist()# Get class namesclass_names_predicted = [class_names[i] for i in class_indices]return class_names_predicted

第九问：
- 连接本地数据库：localhost:3306，用户名：root，密码：root，从数据库poetry中的poems读取前5条数据并打印。
- 把这5条数据导出到Excel
- 发送到钉钉

import java.sql.*;public class Database {public static void main(String[] args) {try {// Connect to local databaseString url = "jdbc:mysql://localhost:3306/poetry";String user = "root";String password = "root";Connection connection = DriverManager.getConnection(url, user, password);// Read first 5 rows from poems tableStatement statement = connection.createStatement();ResultSet resultSet = statement.executeQuery("SELECT * FROM poems LIMIT 5");// Print resultswhile (resultSet.next()) {System.out.println(resultSet.getString("title") + " by " + resultSet.getString("author") + " " + resultSet.getString("content") + " " + resultSet.getString("style") + " " + resultSet.getString("dynasty") + " " + resultSet.getString("notes"));}// Export results to ExcelResultSetToExcelConverter converter = new ResultSetToExcelConverter(resultSet);converter.export("poems.xlsx");DingTalkClient client = new DefaultDingTalkClient("https://oapi.dingtalk.com/robot/send?access_token=ACCESS_TOKEN");OapiRobotSendRequest request = new OapiRobotSendRequest();request.setMsgtype("text");OapiRobotSendRequest.Text text = new OapiRobotSendRequest.Text();text.setContent("Results from database query: " + resultSet.toString());request.setText(text);client.execute(request);// Close connectionresultSet.close();statement.close();connection.close();} catch (SQLException e) {System.out.println("Error connecting to database: " + e.getMessage());}}
}

第十问（源自于实际的物联网开发场景）：
- 编码实现一个OPC UA客户端

import org.eclipse.milo.opcua.sdk.client.OpcUaClient;
import org.eclipse.milo.opcua.stack.core.types.builtin.NodeId;
import org.eclipse.milo.opcua.stack.core.types.builtin.QualifiedName;
import org.eclipse.milo.opcua.stack.core.types.builtin.Variant;
import org.eclipse.milo.opcua.stack.core.types.structured.ReadValueId;
import org.eclipse.milo.opcua.stack.core.types.structured.ReadValueId;
import org.eclipse.milo.opcua.stack.core.types.structured.ReadResponse;
import org.eclipse.milo.opcua.stack.core.types.structured.WriteValue;
import org.eclipse.milo.opcua.stack.core.types.structured.WriteResponse;import java.util.concurrent.CompletableFuture;public class OPCClient {public static void main(String[] args) throws Exception {// Create a new OPC UA clientOpcUaClient client = OpcUaClient.create("opc.tcp://localhost:12686/example");// Connect to the serverclient.connect().get();// Read a value from the serverNodeId nodeId = new NodeId(2, "/HelloWorld/ScalarTypes/Int32");ReadValueId readValueId = new ReadValueId(nodeId, QualifiedName.DEFAULT, null, null);CompletableFuture<ReadResponse> readResponse = client.read(readValueId);Variant value = readResponse.get().getResults()[0].getValue().getValue();System.out.println("Value: " + value);// Write a value to the serverWriteValue writeValue = new WriteValue(nodeId, QualifiedName.DEFAULT, new Variant(42));CompletableFuture<WriteResponse> writeResponse = client.write(writeValue);Boolean success = writeResponse.get().getResults()[0].isGood();System.out.println("Write success: " + success);// Disconnect from the serverclient.disconnect().get();}
}

代码解释

public static int test(int n) {int cnt = 0;while(0 != n) {n = n & (n - 1);++cnt;}return cnt;}public static void main(String[] args) {System.out.println(test(15));}
}

先将上述代码写入一个.java文件，然后 Ctrl+K 输入以下问题。

第十一问：
- 解释下这段代码（或者：这段代码是什么意思？）

// 这段代码是一个Java程序，其中定义了一个名为test的静态方法，该方法接受一个整数参数n并返回一个整数。该方法使用while循环计算n的二进制表示中1的个数，并将其存储在cnt变量中。最后，该方法返回cnt的值。在main方法中，该程序调用test方法并将15作为参数传递给它，然后将结果打印到控制台。

代码优化

public class Improve {public static int test(int n) {int cnt = 0;while(0 != n) {if((n & 1) != 0) {++cnt;n >>= 1;				} else {n >>= 1;				}			}return cnt;	}public static void main(String[] args) {System.out.println(test(15));}
}

先将上述代码写入一个.java文件，然后选择需要优化的代码段或者全选， Ctrl+K 输入以下问题。

第十一问：
- 优化下

public class Improve {public static int test(int n) {int cnt = 0;cnt = Integer.bitCount(n);return cnt;}public static void main(String[] args) {System.out.println(test(15));}
}