I am trying to parse the following String to Byte.But it gives me NumberFormat Exception.Can some body tell me what is the solution for this?
Byte.parseByte("11111111111111111111111110000001", 2);
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2Thats way larger than a byte.Perception– Perception2013-02-18 13:08:31 +00:00Commented Feb 18, 2013 at 13:08
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A byte is made of 8 bits... each of your 0/1 chars is a bit.pcalcao– pcalcao2013-02-18 13:10:27 +00:00Commented Feb 18, 2013 at 13:10
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DV - check javadocs before posting question.djechlin– djechlin2013-02-18 13:13:17 +00:00Commented Feb 18, 2013 at 13:13
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and here's a dup. stackoverflow.com/questions/6996707/java-byte-parsebyte-errordjechlin– djechlin2013-02-18 13:14:36 +00:00Commented Feb 18, 2013 at 13:14
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If you still want to ignore overflows your option is to parse as int and then cast int to byte.Leri– Leri2013-02-18 13:14:37 +00:00Commented Feb 18, 2013 at 13:14
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4 Answers
Byte.parseByte() handles binary string as sign-magnitude not as a 2's complement, so the longest length you can have for a byte is 7 bits with a sign.
In other words, to represent -127, you should use:
Byte.parseByte("-111111", 2);
The following throws NumberFormatException:
Byte.parseByte("10000000", 2);
However, the binary literal of -127 is:
byte b = (byte) 0b10000000;
The same behavior is applied to the other parseXXX() methods.
Comments
Out of range of byte ie -128 to 127. From parseByte(String s,int radix) javadoc:
public static byte parseByte(String s, int radix)throws NumberFormatExceptionParses the string argument as a signed byte in the radix specified by the second argument. The characters in the string must all be digits, of the specified radix (as determined by whether Character.digit(char, int) returns a nonnegative value) except that the first character may be an ASCII minus sign '-' ('\u002D') to indicate a negative value. The resulting byte value is returned. 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') provided that the string is longer than length 1.
- The value represented by the string is not a value of type byte.
Returns: the byte value represented by the string argument in the specified radix Throws: NumberFormatException - If the string does not contain a parsable byte.
4 Comments
Eng.Fouad
This is not correct.
parseXXX() methods handle binary string as sign-magnitude not as a 2's complement. See my answer.
jlordo
@Eng.Fouad: The answer here is correct, actually it's just copy/paste from javadoc without a link to it.
Eng.Fouad
@jlordo I know and that is why I did not down-vote this answer. But they said
Parses the string argument as a signed byte which IMHO is not quite true if radix = 2.jlordo
@Eng.Fouad: Yes, it's true. Java doesn't know unsigned bytes, so they say it parses the string argument as a signed byte. They nowhere say the first digit is the sign bit.
from javadocs
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') provided that the string is longer than length 1.
- The value represented by the string is not a value of type byte.
Your value is the second case that is out of range -128 to 127
Comments
Value is too large to be parsed in byte Try this:
new BigInteger("011111111111111111111111110000001", 2).longValue();
