问题描述
我有此堆栈跟踪(的一部分)
I have this stack trace (part of)
Servlet.service() for servlet action threw exception
java.lang.NumberFormatException: For input string: "37648"
at java.lang.NumberFormatException.forInputString(NumberFormatException.java:48)
at java.lang.Long.parseLong(Long.java:403)
at java.lang.Long.valueOf(Long.java:482)
at java.lang.Long.decode(Long.java:593)
在我的一个日志文件中我不知道什么是真正的输入字符串.但是用户已经进行了相同的堆栈跟踪.
in one of my logfileI don't know what was real input string.But the user had made happen the same stack trace.
这样的堆栈跟踪如何发生?
How such a stacktrace can happen?
推荐答案
可能是因为它们的输入中有前导零.
Probably because they have a leading zero in their input.
运行正常:
public class DecodeLong
{
public static final void main(String[] params)
{
long l;
l = Long.decode("37648");
System.out.println("l = " + l);
}
}
但是,如果您更改此设置:
But if you change this:
l = Long.decode("37648");
对此:
l = Long.decode("037648");
...它变成无效的八进制,并且 Long.parseLong
的异常不包括前导零:
...it becomes invalid octal, and the exception from Long.parseLong
doesn't include the leading zero:
Exception in thread "main" java.lang.NumberFormatException: For input string: "37648"
at java.lang.NumberFormatException.forInputString(Unknown Source)
at java.lang.Long.parseLong(Unknown Source)
at java.lang.Long.valueOf(Unknown Source)
at java.lang.Long.decode(Unknown Source)
at DecodeLong.main(DecodeLong.java:24)
它不包含它,因为 decode
调用 parseLong
不带零,但基数设置为8.
It doesn't include it because decode
calls parseLong
without the zero, but with the base set to 8.
谈论晦涩难懂的话题.:-)因此,如果您通过显示 actual 输入来更新程序以处理异常,则可能会发现这些内容.
Talk about obscure. :-) So if you update your program to handle the exception by showing the actual input, you'll probably find it's something along those lines.
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