Integer.parseInt failing 'For input string: "5000"'

Question

I'll admit it, I'm stumped. It's not a double. It's not outside of the range of an integer. It's not NAN. It's not a non-integer in any way shape or form as far as I can tell.

Why would I get that error?

Here's the code that causes it:

String filename = "confA.txt";

//Make a new filereader to read in confA
FileReader fileReader = new FileReader(filename);

//Wrap into a bufferedReader for sanity's sake
BufferedReader bufferedReader = new BufferedReader(fileReader);

//Get the port number that B is listening to
int portNum = Integer.parseInt(bufferedReader.readLine());

It fails on that last line, stating:

java.lang.NumberFormatException: For input string: "5000"

Which is the number I want.

I've also attempted

Integer portNum = Integer.parseInt(bufferedReader.readLine());

But that didn't work either. Neither did valueOf().

Try printing Arrays.toString(line.toCharArray()) (where line is what you read from the reader). You might have non-printable characters in the string. — Andy Turner
– Andy Turner, Commented Oct 8, 2017 at 19:01
@AndyTurner This appears to be it...somehow. [, 5, 0, 0, 0] was printed. — Tyler Shellberg
– Tyler Shellberg, Commented Oct 8, 2017 at 19:03
Thank you for your help, it's greatly appreciated. It looks like somehow the original text file just had something wrong with it, even if I couldn't see it. I re-wrote it and it works great now. :) — Tyler Shellberg
– Tyler Shellberg, Commented Oct 8, 2017 at 19:09
No problem. It's a simple debugging trick which helps surprisingly often. — Andy Turner
– Andy Turner, Commented Oct 8, 2017 at 19:15

Aleh Maksimovich · Accepted Answer · 2017-10-08 20:21:08Z

Most probably there is some unprintable character somewhere in your file line. Please consider the following example (this was tested in Java 9 jshell)

jshell> String value = "5000\u0007";
value ==> "5000\007"

jshell> Integer.parseInt(value);
|  java.lang.NumberFormatException thrown: For input string: "5000"
|        at NumberFormatException.forInputString (NumberFormatException.java:65)
|        at Integer.parseInt (Integer.java:652)
|        at Integer.parseInt (Integer.java:770)
|        at (#15:1)

Here the string contains the "bell" character at the end. It makes parse to fail while it is not printed in exception text. I think you have something similar. The simpliest way to verify this is to check

String line = bufferedReader.readLine();
System.out.println("line length: " + line.length());

The value other than 4 will support my idea.

isogramc · Accepted Answer · 2018-05-16 11:44:49Z

I had the same problem. I was reading in a flat file with the Buffered reader and saving the contents as an ArrayList of type String, but on performing an integer.parse when retrieving a string value from the list, I realised that there was a whole lot of garbage in the string read from the file, as I got a java.lang.NumberFormatException.

This was the method I implemented with the code (called from my main method) to solve the problem:

` // class level

    private static final Pattern numericPattern = Pattern.compile("([0-9]+).([\\\\.]{0,1}[0-9]*)");

    // in main method after reading in the file
    String b = stripNonNumeric(stringvaluefromfile);
    int a = Integer.parseInt(b);

    public static String stripNonNumeric(String number) {

    //System.out.println(number);

    if (number == null || number.isEmpty()) {
        return "0";
    }

    Matcher matcher = numericPattern.matcher(number);

    // strip out all non-numerics
    StringBuffer sb = new StringBuffer("");
    while (matcher.find()) {
        sb.append(matcher.group());
    }

    // make sure there's only one dot
    int prevDot = -1;
    for (int i = sb.length() - 1; i >= 0; i--) {
        if (sb.charAt(i) == '.') {
            if (prevDot > 0) {
                sb.deleteCharAt(prevDot);
            }
            prevDot = i;
        }
    }

    if (sb.length() == 0) {
        sb.append("0");
    }

    return sb.toString();
}`

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