NumberFormat Exception to parse String to Byte?

Question

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);

A byte is made of 8 bits... each of your 0/1 chars is a bit. — pcalcao, Feb 18 '13 at 13:10
and here's a dup. http://stackoverflow.com/questions/6996707/java-byte-parsebyte-error — djechlin, Feb 18 '13 at 13:14
If you still want to ignore overflows your option is to parse as int and then cast int to byte. — Leri, Feb 18 '13 at 13:14
possible duplicate of [Java, Long.parse binary String](http://stackoverflow.com/questions/14926920/java-long-parse-binary-string) — jlordo, Feb 18 '13 at 13:15

score 2 · Answer 1 · edited Feb 18 '13 at 13:53

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 NumberFormatException
Parses 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.

This is not correct. `parseXXX()` methods handle binary string as sign-magnitude not as a 2's complement. See my answer. — Eng.Fouad, Feb 18 '13 at 13:20
@Eng.Fouad: The answer here is correct, actually it's just copy/paste from javadoc without a link to it. — jlordo, Feb 18 '13 at 13:29
@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. — Eng.Fouad, Feb 18 '13 at 13:32
@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. — jlordo, Feb 18 '13 at 13:36

Eng.Fouad · Answer 2 · 2013-02-18T13:23:36.527

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.

score 1 · Answer 3 · edited Feb 18 '13 at 13:54

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

score 0 · Answer 4 · answered Feb 18 '13 at 13:12

0

Value is too large to be parsed in byte Try this:

new BigInteger("011111111111111111111111110000001", 2).longValue();

answered Feb 18 '13 at 13:12

Raman

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NumberFormat Exception to parse String to Byte?

4 Answers4