Java, Long.parse binary String

Question

Why does this code throw a NumberFormatException :

String binStr = \"1000000000000000000000000000000000000000000000000000000000000000\";
System         


        

Answer 1

Because it's out of range. 1000...000 is 2⁶³, but Long only goes up to 2⁶³ - 1.

Answer 2

This is the same for all of Long, Integer, Short and Byte. I'll explain with a Byte example because it's readable:

System.out.println(Byte.MIN_VALUE); // -128
System.out.println(Byte.MAX_VALUE); // 127
String positive =  "1000000"; // 8 binary digits, +128 
String negative = "-1000000"; // 8 binary digits, -128
String plus     = "+1000000"; // 8 binary digits, +128
Byte.parseByte(positive, 2); //will fail because it's bigger than Byte.MAX_VALUE 
Byte.parseByte(negative, 2); //won't fail. It will return Byte.MIN_VALUE
Byte.parseByte(plus, 2);     //will fail because its bigger than Byte.MAX_VALUE

The digits are interpreted unsigned, no matter what radix is provided. If you want a negative value, you have to have the minus sign at the beginning of the String. JavaDoc says:

Parses the string argument as a signed long 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 or an ASCII plus sign '+' ('\u002B') to indicate a positive value. The resulting long value is returned.

In order to get MAX_VALUE we need:

String max  =  "1111111"; // 7 binary digits, +127 
// or
String max2 = "+1111111"; // 7 binary digits, +127 

Answer 3

This is the largest possible long (9223372036854775807 = 2 exp 63 - 1) in binary format. Note the L at the end of the last digit.

 long largestLong = 0B0111_1111_1111_1111_1111_1111_1111_1111_1111_1111_1111_1111_1111_1111_1111_1111L;

Answer 4

1000000000000000000000000000000000000000000000000000000000000000 is larger than Long.MAX_VALUE.

See https://stackoverflow.com/a/8888969/597657

Consider using BigInteger(String val, int radix) instead.

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EDIT:

OK, this is new for me. It appears that Integer.parseInt(binaryIntegerString, 2) and Long.parseLong(binaryLongString, 2) parse binary as sign-magnitude not as a 2's-complement.

Answer 5

Largest long value is actually:

0111111111111111111111111111111111111111111111111111111111111111b = 9223372036854775807

Answer 6

This is because Long.parseLong cannot parse two's complement representation. The only way to parse two's complement binary string representation in Java SE is BigInteger:

long l = new BigInteger("1000000000000000000000000000000000000000000000000000000000000000", 2).longValue()

this gives expected -9223372036854775808result