Java ParseInt Sanity Check

Question

I'm trying to parse a int from a String array element. Here is my code:

String length = messageContents[k].replace("Content-Length:", "").replace(" ", "");
System.out.println("Length is: " + length);
int test= Integer.parseInt(length);

The System.out.println returns the following: Length is: 23

However, when I try to parse the String into an int, a java.lang.NumberFormatException gets thrown;

java.lang.NumberFormatException: For input string: "23"

I'm a bit confused how 23 wont get parsed into an int. I can only assume that there is some other character in there that is preventing it, but I can't see it for the life of me.

Any suggestions?

Thanks

Update

Despite the String length having only two characters, Java reports its length as three:
Length is: '23'
Length of length variable is: 3
length.getBytes = [B@126804e

I notice that you have a trailing "." on the first line. Perhaps a copy-past error? — Andrew White, Jan 16 '12 at 13:23
Ahh yes, ignore that. Copy and paste error indeed. Thanks :) — Tony, Jan 16 '12 at 13:24
*"I can only assume that there is some other character in there that is preventing it, but I can't see it for the life of me."* 1) Enclose single quotes around the value in the `println` statement. 2) `System.out.println("How long is this piece of String? " + length.length());` — Andrew Thompson, Jan 16 '12 at 13:27
Thanks for that Andrew. Length is: 23 Length of length is: 3 So where the heck is the extra character...Its not appearing in the output? — Tony, Jan 16 '12 at 13:29
What do you get for length.getBytes()? I expect (byte[]) [50, 51] — Wolfgang Kuehn, Jan 16 '12 at 13:30
`System.out.println(Arrays.asList(length.getBytes()));` of course — Donal Fellows, Jan 16 '12 at 13:37

score 7 · Accepted Answer · answered Jan 16 '12 at 13:38

7

Try this variant:

int test= Integer.parseInt(length.trim());

answered Jan 16 '12 at 13:38

Andrew Thompson

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That sorted it, thanks very much! I'm assuming the byte 13 was counted as whitespace by the trim? – Tony Jan 16 '12 at 13:46
Not sure. But `trim()` will get rid of most things in a `String` that are not visible. – Andrew Thompson Jan 17 '12 at 07:40

score 3 · Answer 2 · answered Jan 16 '12 at 13:29

3

There might be unseen characters in this string.

My idea: use a regex with a Pattern/Matcher to remove all the non-numerals in your string, then parse it.

answered Jan 16 '12 at 13:29

Silver Quettier

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Normally I would. But this problem has got me very interested in finding out why the extra character, whatever it may be, isn't appearing within the System.Out. – Tony Jan 16 '12 at 13:37
1

It could be one of those non-printable control characters that you can only find by issuing `System.out.println(Arrays.toString(length.getBytes()));` – anubhava Jan 16 '12 at 13:40
[50, 51, 13] - I think 13 is a carriage return? – Tony Jan 16 '12 at 13:43
1

13 is indeed the carriage return / line feed or `\n` character. – anubhava Jan 16 '12 at 13:50

Java ParseInt Sanity Check

2 Answers2