How to cast to Number from String in java without knowing which number type is present in the String?

huangapple go评论74阅读模式
英文:

How to cast to Number from String in java without knowing which number type is present in the String?

问题

I am writing a compiler, that reads from an input file, parses it and creates various kinds of tokens. Further, in parsing, upon getting a NumLitToken, I retrieve its number value stored as a String and want to save it as a Number for further stages of transformations.

I am not aware whether the String contains int/ float / long / double etc. so I am using NumberFormat.getInstance().parse(x) method and expecting an appropriate casted value, but I don't know why I am not getting a cast to Integer for int values.

Also, if there is any other way better to cast to Number from String, please enlighten me about it.

A small extract:

import java.text.*;
public class Main
{
    public static void main(String[] args) throws ParseException{
        String x = "100";
        Number o  = NumberFormat.getInstance().parse(x);
        System.out.println(o.getClass().toString());
        if(o instanceof Integer){
            System.out.println("int");
        }
    }
}

Output:

class java.lang.Long

UPDATE: Turns out, the method only returns long or double. What to do to get an appropriate cast to Number? Is there a better way rather than trying to cast for every number type?

英文:

<p>

I am writing a compiler, that reads from an input file, parses it and creates various kind of tokens. Further, in parsing, upon getting a NumLitToken , I retrieve its's number value stored as String and want to save it as a Number for further stages of transformations. <br> <br>
I am not aware whether the String contains int/ float / long / double etc. so i am using NumberFormat.getInstance().parse(x) method and expecting appropriate casted value, but i don't know why i am not getting a cast to Integer for int values.<br>
Also, if there is any other way better to cast to Number from String, please enlighten me about it.

</p>

A small extract:

import java.text.*;
public class Main
{
	public static void main(String[] args) throws ParseException{
		String x = &quot;100&quot;;
		Number o  = NumberFormat.getInstance().parse(x);
		System.out.println(o.getClass().toString());
		if(o instanceof Integer){
		    System.out.println(&quot;int&quot;);
		}
	}
}

Output:

class java.lang.Long

<hr>
<p>

UPDATE : Turns out, the method only returns long or double. What to do to get appropriate cast to Number ? Is there a better way rather than trying to cast for every number type?

</p>

答案1

得分: 2

因为文档如此说明:

> 如果可能,返回一个 Long,否则返回一个 Double。

就是这么简单。

如果你只是想解析一个整数,int x = Integer.parseInt("500"); 就能完成任务!

英文:

Because the docs say so:

> Returns a Long is possible, otherwise a Double

Simple as that.

If you're just trying to parse an int, int x = Integer.parseInt(&quot;500&quot;); does the job!

答案2

得分: 1

如在不同的回答中所提到的,这就是它的工作方式。

如果你想要一个根据其大小返回最合适类型的函数,apache-commons中的NumberUtils.createNumber可以为你完成。

英文:

As mentioned in a different answer it is how it works.

If you want to a function that returns the most appropriate type depending its size the apache-commons NumberUtils.createNumber will do that for you.

huangapple
  • 本文由 发表于 2020年10月27日 19:01:11
  • 转载请务必保留本文链接:https://go.coder-hub.com/64553002.html
匿名

发表评论

匿名网友

:?: :razz: :sad: :evil: :!: :smile: :oops: :grin: :eek: :shock: :???: :cool: :lol: :mad: :twisted: :roll: :wink: :idea: :arrow: :neutral: :cry: :mrgreen:

确定