NumberFormatException when parsing strings containing non-integers

Question

I get: java.lang.NumberFormatException.forInputString(NumberFormatException.java:48)

when I try to use:

Integer.parseInt("h");

or

Integer.parseInt("h".toString());

but I was expecting it to work like atoi in c.

What am I doing wrong here?

you are trying to convert a `String` (not a number represent as a String, let say "1", but a letter) to an `int`. That's your mistake, but as already asked: what are you expecting exactly? — ThanksForAllTheFish, Jan 24 '13 at 13:34
`atoi` works almost the same as `Integer.parseInt`, except that instead of throwing an exception it reports an error through a `0` return value. Really, if that's what you're after, just catch the exception and return `0` yourself. Although that's a fairly dumb solution, as an exception is much more appropriate here. — Mattias Buelens, Jan 24 '13 at 13:38
as far i remember (but i'm not expert on math) 'h' is not a number. — user902383, Jan 24 '13 at 13:39
Ahh. Then you need to wrap it in some kind of try catch. Check this question: http://stackoverflow.com/questions/1486077/java-good-way-to-encapsulate-integer-parseint Though I should add that I agree with Mattias...if you're passing in an unparsable string, that's a problem, and it should probably throw an exception, rather than just swallow it and return an arbitrary and potentially misleading value. — Beska, Jan 24 '13 at 14:01

Reimeus · Answer 1 · 2013-01-24T17:47:21.690

3

The String literal "h" obviously does not represent a valid integer.

Assuming h is your String variable, you don't need to use quotes:

String h = "1234";
Integer.parseInt(h);

Update: The function atoi returns an integer value of the input argument and returns zero if the input value is not valid. In Java, you could write this:

public static int javaAToI(String text) {
   try {
      return Integer.parseInt(text);
   } catch (NumberFormatException e) {
      return 0;
   }
}

edited Jan 24 '13 at 17:47

answered Jan 24 '13 at 13:28

Reimeus

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This works if my string is a integer like in your example, but I need to convert text to int. for example string h = "h"; – menemenemu Jan 24 '13 at 13:34
So what value would your "h" have? – bjedrzejewski Jan 24 '13 at 13:35
Are you looking for the ascii value of "h" then? – Reimeus Jan 24 '13 at 13:35
`"h".charAt(0)` will be `'h'` - did you forget the cast to `int` ? ;) – jlordo Jan 24 '13 at 13:39
@Reimeus yes ascii to int – menemenemu Jan 24 '13 at 13:43

score 1 · Answer 2 · edited May 23 '17 at 11:49

This will only work if you parse it a String that resembles Integer.

You are parsing it a literal "h" which can not be interpreted as Integer. If you parse it "123" it will work.

Documentation of your method can be found here and documentation is usually a good place to start your search.

As someont else also suggested- if h is your variable, parse it in this way: Integer.parseInt(h);.

Edit:

If you want to convert a char into int, here is your answer: get char value in java

score 0 · Answer 3 · answered Jan 24 '13 at 13:55

Java is doing exactly what it is supposed to here. In atoi, when you are returned a 0, how do you know if your string contained a 0, or if it had a string that was not parsable? Java handles this by letting you know that you've tried to parse a string that doesn't contain a parsable value...ideally so you can go back and fix the code that is passing in an unparsable value.

In your case, if you wanted to handle the situation exactly like C does, you could just return a 0 if you catch a NumberFormatException. (It would probably be better if you could go back and ensure that you're not passing in bad strings, but that's not always possible.)

NumberFormatException when parsing strings containing non-integers

3 Answers3