Java以指数表示法解析一个数字
声明：本页面是StackOverFlow热门问题的中英对照翻译，遵循CC BY-SA 4.0协议，如果您需要使用它，必须同样遵循CC BY-SA许可，注明原文地址和作者信息，同时你必须将它归于原作者（不是我）：StackOverFlow
原文地址: http://stackoverflow.com/questions/2722122/
Warning: these are provided under cc-by-sa 4.0 license. You are free to use/share it, But you must attribute it to the original authors (not me):
StackOverFlow
Java parse a number in exponential notation
提问by bragboy
I am trying to do a conversion of a String to integer for which I get a NumberFormatException. The reason is pretty obvious. But I need a workaround here. Following is the sample code.
我正在尝试将字符串转换为整数，我得到一个NumberFormatException. 原因很明显。但我需要一个解决方法。以下是示例代码。
public class NumberFormatTest {
public static void main(String[] args) {
String num = "9.18E+09";
try{
long val = Long.valueOf(num);
}catch(NumberFormatException ne){
//Try to convert the value to 9180000000 here
}
}
}
I need the logic that goes in the comment section, a generic one would be nice. Thanks.
我需要评论部分的逻辑，通用的逻辑会很好。谢谢。
采纳答案by Joachim Sauer
Use Double.valueOf()and cast the result to long:
使用Double.valueOf()并将结果转换为long：
Double.valueOf("9.18E+09").longValue()
回答by Bozho
BigDecimal bd = new BigDecimal("9.18E+09");
long val = bd.longValue();
But this adds overhead, which is not needed with smaller numbers. For numbers that are representable in long, use Joachim Sauer's solution.
但这会增加开销，这对于较小的数字是不需要的。对于可在 中表示的数字long，请使用 Joachim Sauer 的解决方案。
