why Integer.parseInt(“11111111111111111111111111111111”,2) throws exception in java? [closed]

问题

why does Integer.parseInt("11111111111111111111111111111111",2) throw

java.lang.NumberFormatException: For input string: "11111111111111111111111111111111"

In java integer is 32 bit, I expect a valid return value, what is going wrong here ?

回答1:

Integer is truly 32 bit, but one bit is used for positive/negative sign.

This code

// 31 instead of 32
System.out.println(Integer.parseInt("1111111111111111111111111111111",2)); 
System.out.println(Integer.MAX_VALUE);

will produce exactly same number 2147483647.

EDIT: Integer.parseInt specification states that the correct way to specify negative value is to use - minus sign:

/**
 * Parses the string argument as a signed integer 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 {@link java.lang.Character#digit(char, int)} returns a 
 * nonnegative value), except that the first character may be an 
 * ASCII minus sign <code>'-'</code> (<code>'&#92;u002D'</code>) to 
 * indicate a negative value. The resulting integer value is returned. 
 * <p>
 * An exception of type <code>NumberFormatException</code> is
 * thrown if any of the following situations occurs:
 * <ul>
 * <li>The first argument is <code>null</code> or is a string of
 * length zero.
 * <li>The radix is either smaller than 
 * {@link java.lang.Character#MIN_RADIX} or
 * larger than {@link java.lang.Character#MAX_RADIX}. 
 * <li>Any character of the string is not a digit of the specified
 * radix, except that the first character may be a minus sign
 * <code>'-'</code> (<code>'&#92;u002D'</code>) provided that the
 * string is longer than length 1.
 * <li>The value represented by the string is not a value of type
 * <code>int</code>. 
 * </ul><p>
 * Examples:
 * <blockquote><pre>
 * parseInt("0", 10) returns 0
 * parseInt("473", 10) returns 473
 * parseInt("-0", 10) returns 0
 * parseInt("-FF", 16) returns -255
 * parseInt("1100110", 2) returns 102
 * parseInt("2147483647", 10) returns 2147483647
 * parseInt("-2147483648", 10) returns -2147483648
 * parseInt("2147483648", 10) throws a NumberFormatException
 * parseInt("99", 8) throws a NumberFormatException
 * parseInt("Kona", 10) throws a NumberFormatException
 * parseInt("Kona", 27) returns 411787
 * </pre></blockquote>
 *
 * @param      s   the <code>String</code> containing the integer 
 *          representation to be parsed
 * @param      radix   the radix to be used while parsing <code>s</code>.
 * @return     the integer represented by the string argument in the
 *             specified radix.
 * @exception  NumberFormatException if the <code>String</code>
 *         does not contain a parsable <code>int</code>.
 */

回答2:

You assume that you are passing the raw bits to the method, but in fact you are passing a binary representation of the number. Therefore you can specify up to 31 bit, and a sign. This gives the result you have expected:

Integer.parseInt("-1", 2);

The contract of the method is such that you can specify any (reasonable) base, for example the hexatridecimal base:

Integer.parseInt("-1", 36);

From this it should be obvious that this is about the conversion from a representation in an arbitrary number system to an int, as opposed to passing raw contents of the int.

回答3:

The integer is too big. The maximum value is 2 147 483 647

回答4:

Integers in java are signed integer. Thus the first bit is needed for the - or + information.

As the java documentation of Integer.MAX_VALUE says:

Integer.MAX_VALUE A constant holding the maximum value an int can have, 2^31-1.

The 32 bits "11111111111111111111111111111111" would be 4294967296 and therefore be out of the range of an integer.

回答5:

Max value of int32 is 2,147,483,647

来源：https://stackoverflow.com/questions/18170101/why-integer-parseint11111111111111111111111111111111-2-throws-exception-in-j

标签

java

integer