go gpu 编程（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）

时间2025-06-17 03:44:59分类IT科技浏览5610

导读：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...

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 theres 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 theres 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 Algorithm import numpy as np import matplotlib.pyplot as plt class KMeans: def __init__(self, k=2, max_iter=100): self.k = k self.max_iter = max_iter def fit(self, X): self.centroids = {} # Initialize centroids randomly for i in range(self.k): self.centroids[i] = X[np.random.choice(len(X))] # Iterate until convergence or max iterations reached for i in range(self.max_iter): self.clusters = {} for j in range(self.k): self.clusters[j] = [] # Assign each data point to the closest centroid for 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 points for 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 order for (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)); // Intersection Set<String> intersection = new HashSet<>(set1); intersection.retainAll(set2); System.out.println("Intersection: " + intersection); // Union Set<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 URL response = requests.get(image_url) img = Image.open(BytesIO(response.content)) # Predict image results = model(img) # Get class predictions class_predictions = results.pred[0].softmax(-1) # Get class indices class_indices = class_predictions.argmax(-1).tolist() # Get class names class_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 database String 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 table Statement statement = connection.createStatement(); ResultSet resultSet = statement.executeQuery("SELECT * FROM poems LIMIT 5"); // Print results while (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 Excel ResultSetToExcelConverter 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 connection resultSet.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 client OpcUaClient client = OpcUaClient.create("opc.tcp://localhost:12686/example"); // Connect to the server client.connect().get(); // Read a value from the server NodeId 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 server WriteValue 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 server client.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)); } }

从结果看， Cursor 将我们自己编写的通过位运算右移计算二进制表示中1的个数，改成了一个内置方法调用。

小总结

其实，除了上述的入门体验，还有其他很多事情可以做，eg: 编写测试用例，解决力扣编程问题，网络爬虫，制作网页，小游戏编程，you name it ，这一切仅受限于我们的想象力。

以下引用池建强的一句话：

人们需要警惕的是，当年因为汽车的诞生而失业的马车夫，他们并不是转行去干司机了，而是真失业了，或者去干别的苦力活。

当上时代司机的，是另一群人。

If you have any questions or any bugs are found, please feel free to contact me.

Your comments and suggestions are welcome!