为什么我的应用程序在执行此代码时崩溃? [英] Why my app crashing executing this code?

问题描述

Android. Java.Lang.NullPointerException执行以下代码:

 尝试 {
//  sc.setReceiveBufferSize(163840); 
sc = 新套接字("  span>， 8090 )；
SocketAddress sa = 新 InetSocketAddress(" ， 8090 )；
sc.connect(sa);
   } 捕获(异常x)
   {
       logn.setText(x.toString()); //  Java.Lang.NullPointerException，为什么?

   } 

登录在OnCreate函数中声明，并且我写了"EditText登录；"在班上
我已经注释了"sc.setReceiveBufferSize(163840);"，因此现在应用程序崩溃了，没有任何异常.观点.查看您的代码，找出原因.如果您对运行时行为的关注最小，请始终使用调试器.

—SA

第二个谢尔盖的回答是，套接字连接失败.

您还应该这样写:

 尝试 {
  sc.setReceiveBufferSize( 163840 );
  sc = 新套接字("  span>， 8090 )；
  SocketAddress sa = 新 InetSocketAddress(" ， 8090 )；
  sc.connect(sa);
}
捕获(异常x){
       System.out.println(x.printStackTrace()); // 以获取完整的异常
   } 

我还建议对变量进行更好的命名-您将不熟悉sc/sa/s ....将它们命名为"socket"和"address"，依此类推.

Android. Java.Lang.NullPointerException execiting this code:

try{
//sc.setReceiveBufferSize(163840);
sc = new Socket("46.247.186.175", 8090);
SocketAddress sa = new InetSocketAddress("46.247.186.175", 8090);
sc.connect(sa);
   }catch(Exception x)
   {
       logn.setText(x.toString());//Java.Lang.NullPointerException, Why?

   }

EDIT: logn is declared in OnCreate func and I wrote "EditText logn;" in the Class
EDIT2: I have commented "sc.setReceiveBufferSize(163840);", so now app crashing without any exceptions.

解决方案

Apparently, because logn is null at this point. Look at your code and find out why. Should you have a smallest concern about your runtime behavior, always use the debugger.

—SA

I second Sergey''s answer, the Socket Connection is failing.

You should also write that different:

try{
  sc.setReceiveBufferSize(163840);
  sc = new Socket("46.247.186.175", 8090);
  SocketAddress sa = new InetSocketAddress("46.247.186.175", 8090);
  sc.connect(sa);
}
catch(Exception x) {
       System.out.println(x.printStackTrace()); // to get the full exception
   }

I also recommend better naming of variables - you will loose track with sc/sa/s... . Name them "socket" and "address" and so on.

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