I want to parse an String containing 8 hex-digits (4bytes) but i got an NumberFormatException. What is wrong here?
assertThat(Integer.parseInt("FFFF4C6A",16),is(0xFFFF4C6A));
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5 Answers
Your number represents a number greater than that assignable to an int. Try:
Long.parseLong("FFFF4C6A", 16);
which gives 4294921322.
From the doc:
An exception of type
NumberFormatExceptionis thrown if any of the following situations occurs:
- The first argument is null or is a string of length zero.
- The radix is either smaller than
Character.MIN_RADIXor larger thanCharacter.MAX_RADIX.- Any character of the string is not a digit of the specified radix, …
- The value represented by the string is not a value of type int.
and it's the 4th case that you're hitting.
Comments
You've exceeded the range of an integer.
Integer.MAX_VALUE = 2147483647
0xFFFF4C6A = 4294921322
Parsing it as a Long works:
Long.parseLong("FFFF4C6A",16)
1 Comment
Vlad
perhaps 16 instead of
Character.MAX_RADIX expresses the OP's intention betterThat is because the Integer.parseInt("FFFF4C6A",16) provided exceeds Integer.MAX_VALUE which is defined as public static final int MAX_VALUE = 0x7fffffff;
Now, as per the Javadoc for parseInt(...), you would hit a NumberFormatException in either of the following cases:
An exception of type NumberFormatException is thrown if any of the following situations occurs:
- The first argument is null or is a string of length zero.
- The radix is either smaller than Character.MIN_RADIX or larger than Character.MAX_RADIX.
- Any character of the string is not a digit of the specified radix, except that the first character may be a minus sign '-' ('\u002D') or plus sign '+' ('\u002B') provided that the string is longer than length 1.
- The value represented by the string is not a value of type int.
In your case, since the String value supplied exceeds Integer.MAX_VALUE, you're satisfying the 4th clause for NumberFormatException
Possible Solution: In order to parse this, use Long.parseLong(...) where the MAX_VALUE is defined as `public static final long MAX_VALUE = 0x7fffffffffffffffL
Comments
If you just want to represent that hex string as an integer (since it is 32 bits), you need to use BigInteger:
new BigInteger("FFFF4C6A", 16).intValue()
Comments
I don't know the assertThat() method, but your hexadecimal number "FFFF4C6A" is to big for an integer.
For example, if you write :
int number = Integer.parseInt("FFFF4C6A",16)
you'll get the same error. A correct way to write the code would be :
double number = Integer.parseInt("FFFF4C6A",16)
