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I have this binary value : 

11001001111001010010010010011010.

In java 7, if I declare:

int value = 0b11001001111001010010010010011010;

and print the value I get -907729766.

How can I achieve this in Java 6 as well?

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3 Answers 3

Reset to default
6

You have to parse it as a long first, then narrow it into int.

String s = "11001001111001010010010010011010";
int i = (int)Long.parseLong(s, 2); //2 = binary
System.out.println(i);

Prints:

-907729766
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Comments

1

Just so you know since Java 8 there is Integer.parseUnsignedInt method which lets you parse your data without problems

String s = "11001001111001010010010010011010";
System.out.println(Integer.parseUnsignedInt(s,2));

Output: -907729766

Java is open source so you should be able to find this methods code and adapt it to your needs in Java 6.

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0

You can't use the value you entered, because it is outside of the limit of Integer. Use the radix 2 to convert it on Java 6.

This will fail

    System.out.println(Integer.parseInt("11001001111001010010010010011010", 2));

with Exception in thread "main" java.lang.NumberFormatException: For input string: "11001001111001010010010010011010" at java.lang.NumberFormatException.forInputString(NumberFormatException.java:65) at java.lang.Integer.parseInt(Integer.java:495) at Test.main(Test.java:7)

But this will work

 System.out.println(Long.parseLong("11001001111001010010010010011010", 2));

the output will be 3387237530

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3 Comments

That is because the value int -907729766 is actually the long value 3387237530. Try this: long l = 3387237530L; System.out.println((int)l);
@AlexandreSantos Not exactly. Decimal value of int -907729766 is same as decimal value of long -907729766L. It binary representation of int -907729766 is same as binary representation of long 3387237530L (at least at last 32 bits).
Sure, but System.out.println((int)l) doesn't print the binary, it prints with the overflow.

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