blocks|key|1420380|text|试试这个：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1420381|if+(items.elementAt(1)+instanceof+Double)+{
+++sum.add(+i,+items.elementAt(1));
}|code-block|syntax|javascript|1420382|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|J|8|@]|9|@]|A|$E|F]]|$1|G|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|H|$]]

Try this:

<pre><code>if (items.elementAt(1) instanceof Double) {
 sum.add( i, items.elementAt(1));
}
</code></pre>

blocks|key|1205456|text|反射速度较慢，但适用于想知道它是狗还是猫的类型而不是Animal实例的情况。所以你会做类似这样的事情：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1205457|if(null+!=+items.elementAt(1)+&&+items.elementAt(1).getClass().toString().equals("Cat"))
{
//do+whatever+with+cat..+not+any+other+instance+of+animal..+eg.+hideClaws();
}|code-block|syntax|javascript|1205458|不说上面的答案是不起作用的，除非空检查部分是必要的。|1205459|另一种解决方法是使用泛型，这样你就可以保证拥有两个元素。|1205460|List<Double>+items+=+new+ArrayList<Double>();|1205461|entityMap^0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|O|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|P|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|Q|8|@]|9|@]|A|$]]|$1|I|3|J|5|6|7|R|8|@]|9|@]|A|$]]|$1|K|3|L|5|D|7|S|8|@]|9|@]|A|$E|F]]|$1|M|3|-4|5|6|7|T|8|@]|9|@]|A|$]]]|N|$]]

Reflection is slower, but works for a situation when you want to know whether that is of type Dog or a Cat and not an instance of Animal. So you'd do something like:

<pre><code>if(null != items.elementAt(1) &amp;&amp; items.elementAt(1).getClass().toString().equals("Cat"))
{
//do whatever with cat.. not any other instance of animal.. eg. hideClaws();
}
</code></pre>

Not saying the answer above does not work, except the null checking part is necessary.

Another way to answer that is use generics and you are guaranteed to have Double as any element of items.

<pre><code>List&lt;Double&gt; items = new ArrayList&lt;Double&gt;();
</code></pre>

blocks|key|1420485|text|使用正则表达式来实现此任务。请参考下面的代码。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1420486|public+static+void+main(String[]+args)+{
++++try+{

++++++++BufferedReader+reader+=+new+BufferedReader(new+InputStreamReader(System.in));
++++++++System.out.print("Enter+your+content:+");
++++++++String+data+=+reader.readLine();++++++++++++
++++++++boolean+b1+=+Pattern.matches("%5E\\d%2B$",+data);
++++++++boolean+b2+=+Pattern.matches("[0-9a-zA-Z([%2B-]?\\d*\\.%2B\\d*)]*",+data);+
++++++++boolean+b3+=+Pattern.matches("%5E([%2B-]?\\d*\\.%2B\\d*)$",+data);
++++++++if(b1)+{
++++++++++++System.out.println("It+is+integer.");
++++++++}+else+if(b2)+{
++++++++++++System.out.println("It+is+String.+");
++++++++}+else+if(b3)+{
++++++++++++System.out.println("It+is+Float.+");
++++++++}+++++++++++
++++}+catch+(IOException+ex)+{
++++++++Logger.getLogger(TypeOF.class.getName()).log(Level.SEVERE,+null,+ex);
++++}
}|code-block|syntax|javascript|1420487|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|J|8|@]|9|@]|A|$E|F]]|$1|G|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|H|$]]

Use regular expression to achieve this task. Please refer the below code.

<pre><code>public static void main(String[] args) {
 try {

 BufferedReader reader = new BufferedReader(new InputStreamReader(System.in));
 System.out.print("Enter your content: ");
 String data = reader.readLine(); 
 boolean b1 = Pattern.matches("^\\d+$", data);
 boolean b2 = Pattern.matches("[0-9a-zA-Z([+-]?\\d*\\.+\\d*)]*", data); 
 boolean b3 = Pattern.matches("^([+-]?\\d*\\.+\\d*)$", data);
 if(b1) {
 System.out.println("It is integer.");
 } else if(b2) {
 System.out.println("It is String. ");
 } else if(b3) {
 System.out.println("It is Float. ");
 } 
 } catch (IOException ex) {
 Logger.getLogger(TypeOF.class.getName()).log(Level.SEVERE, null, ex);
 }
}
</code></pre>

blocks|key|1420519|text|由于这是来自谷歌的第一个问题，我将在这里添加JavaScript风格的typeof替代方案：|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|1420520|myObject.getClass().getName()+//+String|code-block|syntax|javascript|1420521|entityMap^0|Z|6|0|0^^$0|@$1|2|3|4|5|6|7|M|8|@$9|N|A|O|B|C]]|D|@]|E|$]]|$1|F|3|G|5|H|7|P|8|@]|D|@]|E|$I|J]]|$1|K|3|-4|5|6|7|Q|8|@]|D|@]|E|$]]]|L|$]]

Since this is the first question from Google I'll add the JavaScript style <code>typeof</code> alternative here as well:

<pre><code>myObject.getClass().getName() // String
</code></pre>

I'm a Java newbie. I'm trying to figure out whether a number is a Double with something like this:

<pre><code>if ( typeof ( items.elementAt(1) )== Double ) {
 sum.add( i, items.elementAt(1));
}
</code></pre>

Would appreciate if someone could tell me how to rearrange the syntax to make this work properly.

Figuring out whether a number is a Double in Java

Java

我是个Java新手。我正在尝试计算一个数字是否是一个双精度数，如下所示：if ( typeof ( items.elementAt(1) )== Double ) {       sum.add( i, items.elementAt(1));}如果有人能告诉我如何重新安排语法使其正常工作，我将不胜感激。

问判断Java中的数字是否为双精度
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