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Issue with float to binary

Krishna Venkatesan

Joined: Jul 25, 2006
Posts: 1
We need to convert Float to Binary and we are using the following APIs:
1) Float.floatToIntBits(<float>
2) Integer.toBinaryString(<intbits>

The usage is:
String <string> = Integer.toBinaryString(Float.floatToIntBits(<float>

For most cases we get the correct Binary String. However, when the float variable has value = 8 digit integers, we run into the following problem:

Problem: floatToIntBits Returns the SAME output for different Input values.


public class TestFloatToBinary
//Function to convert the float value to binary
public static void main(String args[])
float f1 = 17012219;
float f2 = 17012220;
float f3 = 17012221;
float f4 = 17012222;

String sBinary1 = Integer.toBinaryString(Float.floatToIntBits(f1));
String sBinary2 = Integer.toBinaryString(Float.floatToIntBits(f2));
String sBinary3 = Integer.toBinaryString(Float.floatToIntBits(f3));
String sBinary4 = Integer.toBinaryString(Float.floatToIntBits(f4));

System.out.println("float to intbits for f1 = " + Float.floatToIntBits(f1));
System.out.println("float to intbits for f2 = " + Float.floatToIntBits(f2));
System.out.println("float to intbits for f3 = " + Float.floatToIntBits(f3));
System.out.println("float to intbits for f4 = " + Float.floatToIntBits(f4));

System.out.println("Binary for f1 = " + sBinary1);
System.out.println("Binary for f2 = " + sBinary2);
System.out.println("Binary for f3 = " + sBinary3);
System.out.println("Binary for f4 = " + sBinary4);


C:\work\krishna\java>java TestFloatToBinary
float to intbits for f1 = 1266797310
float to intbits for f2 = 1266797310
float to intbits for f3 = 1266797310
float to intbits for f4 = 1266797311
Binary for f1 = 1001011100000011100101011111110
Binary for f2 = 1001011100000011100101011111110
Binary for f3 = 1001011100000011100101011111110
Binary for f4 = 1001011100000011100101011111111

As seen above, floatToIntBits and hence Integer.toBinaryString return the SAME value for Inputs 17012219, 17012220, 17012221.

What is the reason why this happens?

Are we missing something here??

Appreciate your help!

Thank you.
Jesper de Jong
Java Cowboy
Saloon Keeper

Joined: Aug 16, 2005
Posts: 15093

You need to understand that floating point data types such as float and double do not have infinite precision.

You cannot store any number with an arbitrary number of digits and expect that it is stored with infinite precision.

The precision of a 32-bit IEEE 754 floating point number such as float in Java is about 5 or 6 decimal digits, so it's no surprise that numbers that differ beyond about the 5th digit only look the same in binary form.

IEEE Standard 754 Floating Point Numbers
IEEE floating-point standard (Wikipedia)
[ July 26, 2006: Message edited by: Jesper Young ]

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Campbell Ritchie

Joined: Oct 13, 2005
Posts: 46405
Apart from the precision problem, there is something else not right here. You start with positive numbers, and end up with a String starting with a 1.

If you count the bits printed, you find they only come to 32 if you use a negative number. So your "toBinaryString" is losing its leading zeroes.
Ernest Friedman-Hill
author and iconoclast

Joined: Jul 08, 2003
Posts: 24199

Please don't cross-post. I deleted the copy of this thread in the Advanced forum.

[Jess in Action][AskingGoodQuestions]
Edwin Dalorzo
Ranch Hand

Joined: Dec 31, 2004
Posts: 961
Another thing that may be important to take into account is that all NaN results will produce the same result using Double.doubleToLongBits(). In this case, to diferentiate a NaN from another you might consider using Double.doubleToRawLongBits()

Vlado Zajac
Ranch Hand

Joined: Aug 03, 2004
Posts: 245
Because of limited precision of floating point representation, f1, f2 and f3 have same value.

Try adding

to your example.
I agree. Here's the link:
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