无法将很长的二进制字符串解析为数字类型

Question

Folks, I am trying to parse an extremely long binary String to its decimal equivalent, but its throwing the NumberFormatException .

String s = "1111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111" +
        "1111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111" +
        "1111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111" +
        "1111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111" +
        "1111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111" +
        "1111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111" +
        "1111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111" +
        "1111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111" +
        "1111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111" +
        "1111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111" +
        "1111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111" +
        "1111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111" +
        "1111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111" +
        "1111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111" +
        "111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111" +
        "1111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111" +
        "1111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111" +
        "1111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111" +
        "1111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111" +
        "1111111111111111"; //String length is 1969

long n = Long.parseLong(s, 2); //line no. 25

System.out.println(n);

But it's giving the below-mentioned Runtime error:

Exception in thread "main" java.lang.NumberFormatException: For input string: "1111[...]111" at java.lang.NumberFormatException.forInputString(NumberFormatException.java:65) at java.lang.Long.parseLong(Long.java:592) at Check2.main(Check2.java:25)

Tried using BigInteger,valueOf() methods too, but all efforts are going in vain.

Please do let me know if there's any other way to achieve the desired result.

Answer 1

The java long data type has a minimum value of -9,223,372,036,854,775,808 and a maximum value of 9,223,372,036,854,775,807 . The number you have is much greater than the maximum limit allowed.

You can use BigInteger for your String instead.

BigInteger result = new BigInteger(inputString, 2);

Answer 2

Both int and long have a maximum (and miminum) value, see the documentation for int and long accordingly. The decimal representation of your binary string exceeds these limits.

You might use a BigDecimal instead to parse the String. It provides a constructor accepting a String with the number and the radix to use. As your string is in binary format, you should pass the radix value of 2.

BigInteger myValue = new BigInteger(s, 2);

Answer 3

An answer to the amendment:

If you want to calculate with BigInteger you have to do all operations with BigInteger . In the code fragment

    BigInteger n = new BigInteger(s, 2);
    int count = 0;
    while (n.intValue() != 0) {
        if (n.intValue() % 2 == 0) {
            n = BigInteger.valueOf(n.intValue() / 2);
            count++;    
        } else {
            n = BigInteger.valueOf(n.intValue() - 1);
            count++;
        }

    }
    System.out.println(count);

you loose important information every time you call n.intValue() and every time you create a new BigInteger from an int (actually a long , the public BigInteger.valueOf() only accepts a long .)

This is because an int can only store 32 bits, the value you start with however has 1969 bits. n.intValue() extracts the lowest 32 bits, and after BigInteger.valueOf(n.intValue() - 1) / BigInteger.valueOf(n.intValue() / 2) all but those last 32 bits are lost.

The code works if you replace it with

    BigInteger n = new BigInteger(s, 2);
    int count = 0;
    while (!n.equals(BigInteger.ZERO)) {
        count++;
        if (!n.testBit(0)) {
            n = n.divide(BigInteger.TWO);
        } else {
            n = n.subtract(BigInteger.ONE);
        }
    }
    return count;

---

Why does your code lead to an endless loop?

The endless loop arises from the fact that the int value that has all bits set is -1.

Your original loop produces these values:

1. -1 is odd, therefore it subtracts 1 which gives -2
2. -2 is even, the value is divided by 2 which gives -1

Answer 4

You can use this following method BigInteger() for integer conversion:

BigInteger bi = new BigInteger(inputstring, 2);

or you can use BigDecimal() like so:

BigDecimal bd1 = new BigDecimal(inputstring.charAt(0)=='1'?1:0);
BigDecimal two = new BigDecimal(2);
for (int i = 1; i<inputstring.length(); i++) {
    bd1 = bd1.multiply(two);
    bd1 = bd1.add(new BigDecimal(inputstring.charAt(i)=='1'?1:0));
}
System.out.println("Big decimal number is"+ bd1);