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 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!
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