hotspot

JAVA SE 6 HOTSPOT VM OPTIONS -转

由 99封情书 提交于 2020-01-07 17:19:12
【推荐】2019 Java 开发者跳槽指南.pdf(吐血整理) >>> Platform Default VM Architecture OS client VM if server-class, server VM; otherwise, client VM server VM SPARC 32-bit Solaris X i586 Solaris X Linux X Microsoft Windows X SPARC 64-bit Solaris — X AMD64 Linux — X Microsoft Windows — X 很容易看出，在Windows 里面，启动的永远都是client VM. Java 6 JVM参数选项大全（中文版） 3 人收藏此文章， 收藏此文章 发表于6个月前 , 已有 383 次阅读 共 个评论 3 人收藏此文章 本文 是基于最新的 SUN官方 文档 Java SE 6 Hotspot VM Options 编写的译文。主要介绍 JVM 中的非稳态选项及其使用说明。 为了让读者明白每个选项的含义，作者在原文基础上 补充了大量的资料 。希望这份文档，对正在研究 JVM 参数的朋友有帮助！ 另外，考虑到本文档是初稿，如有描述错误，敬请指正。 非稳态 选项 使用说明 -XX:+<option> 启用选项 -XX:-<option> 不启用选项 -XX:

Mobile Hotspot Name and Password

由 ￣綄美尐妖づ 提交于 2020-01-06 06:34:31
问题 I have to get the name and password of my mobile hotspot programmatically in android studio. How do I do it? WifiManager wifiManager = (WifiManager) getApplicationContext().getSystemService(WIFI_SERVICE); WifiInfo wifiInfo = wifiManager.getConnectionInfo(); Toast.makeText(this,"SSID:"+wifiInfo.getSSID(),Toast.LENGTH_LONG).show(); This code gives me SSID of wifi I am connected to. I need name of my Mobile hotspot. 回答1: You can get the wificonfiguration of your hotspot in API<26 using

Mac OS HSDB HotSpot Debugger can not attach to the process

由 女生的网名这么多〃 提交于 2020-01-06 05:15:29
问题 I want to use HSDB to debug my java code. I follow the steps blow: Debug my code in IntelliJ and stop at a breakpoint Start HSDB: java -cp ,:/Library/Java/JavaVirtualMachines/jdk1.8.0_161.jdk/Contents/Home/lib/sa-jdi.jar sun.jvm.hotspot.HSDB use jps to get the pid attach to the pid, but error occurs. My java version: java version "1.8.0_161" Java(TM) SE Runtime Environment (build 1.8.0_161-b12) Java HotSpot(TM) 64-Bit Server VM (build 25.161-b12, mixed mode) My os version: System Version:

How to turn off wifi hotspot programmatically in Android 8.0 (Oreo) (setWifiApEnabled is not support this version anymore)

由 最后都变了- 提交于 2020-01-04 07:10:49
问题 I reference the code to turn on hotspot in Android 8.0, it is work. But I have no idea about how to disable it @RequiresApi(api = Build.VERSION_CODES.O) private void turnOnHotspot(){ WifiManager manager = (WifiManager) getApplicationContext().getSystemService(Context.WIFI_SERVICE); manager.startLocalOnlyHotspot(new WifiManager.LocalOnlyHotspotCallback(){ @Override public void onStarted(WifiManager.LocalOnlyHotspotReservation reservation) { super.onStarted(reservation); Log.d(TAG, "Wifi

Can HotSpot inline lambda function calls?

由 纵然是瞬间 提交于 2020-01-02 03:11:41
问题 Considering the code: someList.forEach(x -> System.out.format("element %s", x)); Theoretically, it should be possible to inline this code and eliminate the indirect function calls by first inlining the forEach method, and then inlining the lambda function body in the inlined forEach code. Is HotSpot capable of performing this optimization? What restrictions govern whether or not it is performed in a particular situation? 回答1: Your lambda expression is compiled into an ordinary method, while

Android 2.2 wifi hotspot API

由 冷暖自知 提交于 2019-12-28 01:42:51
问题 What is the API call I need to make in Android 2.2 (Froyo) to create a Wifi hotspot (as seen in the Tethering and Portable Hotspot settings item). 回答1: You can call private boolean setWifiApEnabled(WifiConfiguration wifiConfig, boolean enabled); using reflection :) after getting the WifiManager use the reflection to get the WifiManager declared methods, look for this method name setWifiApEnabled and invoke it through the WifiManager object These API are marked as @hide, so currently you

How to open Android Settings Wifi Hotspot Menu Pragmatically in Java?

由 冷暖自知 提交于 2019-12-25 19:02:05
问题 I want to open below or if this is not possible then below settings through java code in android. I know i can open settings using startActivityForResult(new Intent(Settings.ACTION_SETTINGS), 0); but i am not able to find appropriate strings for either of the above menus. Please if anyone can help. 回答1: Try this, startActivity(new Intent(Settings.ACTION_WIFI_SETTINGS)); 来源： https://stackoverflow.com/questions/55083427/how-to-open-android-settings-wifi-hotspot-menu-pragmatically-in-java

Turn on/off WiFi hotspot programmatically

由 早过忘川 提交于 2019-12-24 02:16:26
问题 I need help in creating a C# script that set a android WiFi in hotspot mode. Here is the code that I managed to create. public bool setAPEnabled(bool enabled) { using (AndroidJavaObject activity = new AndroidJavaClass("com.unity3d.player.UnityPlayer").GetStatic<AndroidJavaObject>("currentActivity")) { try { if(isWifiEnabled()==true){ setWifiEnabled(false); } using (var wifiManager = activity.Call<AndroidJavaObject>("getSystemService", "wifi")) { return wifiManager.Call<bool>("setWifiApEnabled

NEHotspotConfigurationManager getting this alert:“Unable to join the network<name of network>” while error is nil

由 筅森魡賤 提交于 2019-12-23 09:29:32
问题 So i am trying to monitor the connection status by closers : func reconnect(success: @escaping () -> Void, failure: @escaping () -> Void) { let manager = NEHotspotConfigurationManager.shared let ssid = CameraManager.camera.uuid let password = "password" let isWEP = false let hotspotConfiguration = NEHotspotConfiguration(ssid: ssid, passphrase: password, isWEP: isWEP) hotspotConfiguration.joinOnce = true manager.apply(hotspotConfiguration) { (error) in if (error != nil) { if let error = error

How to create a hotspot network in iOS app using Swift

由 霸气de小男生 提交于 2019-12-23 02:47:09
问题 I want to create a hotspot network in iOS app using swift. Also please suggest ony apple approved APIs. Thanks! 回答1: NEHotspotHelper is an official API to do exactly this kind of stuff. Though the API is not an 'Open to All' kind. It needs the entitlement 'com.apple.developer.networking.HotspotHelper' to be included in your app. To use this entitlement you need to ask for a permission from Apple to use this class with your app's bundle identifier with a clear reason for why you need it with
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