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javabinarydecimal
Noor_Muhammad
Noor_Muhammad

Reputation: 25

Convert 64-Bit Binary floating-point number into a base 10 decimal

I have a task to convert a 64-bit binary floating-point number into a base 10 decimal. An example here:

Input: 64-bit binary floating-point number: 0100000000001001001000011111101101010100010001000010110100011000

Output: Base 10 Decimal: 3.14

what I know here, first character is a sign = 0 (positive) , 1(negative). Next 11 bits represent the exponent, and the rests are Mantissa.

My problem is I don't know the math equation or the math solution to convert it. I am going to solve this task using Java.

Upvotes: 0

Views: 1567

Answers (3)

WJS
WJS

Reputation: 40034

Here is the explanation. As you already stated, the main parts are:

- sign     = 0
- exponent = 10000000000
- mantissa = 1001001000011111101101010100010001000010110100011000

Since all mantissa values are shifted left (normalized) until the first bit is a 1, that bit can be implied. So the actual mantissa prefixed with a 1, the value would be

- mantissa = 11001001000011111101101010100010001000010110100011000

The base offset for the exponent is 1023. The exponent above is 1024 so the exponent would be 1 + an additional 1 for the implied bit.

.11 x 22 So the whole number part would be binary 11 or decimal 3.

double whole = 3;

The rest of the mantissa is the fractional part and starts where the whole number stopped.

fraction = 001001000011111101101010100010001000010110100011000

so given a string of bits as above, the fractional part can be added to the whole by dividing each successive bit by ever increasing powers of 2. (e.g. 0/2 + 0/4 + 1/8 + etc)

double div = 1.;
for (char c : fract.toCharArray()) {
     div*=2.;
     whole += (c-'0')/div;
}
System.out.println(whole);

prints

3.141592653589793

Note: The above was cheating somewhat in that it used a double to compute the fractional part. There are more involved methods to emulate floating point math but the purpose here was to demonstrate extracting the bit fields to create a double value. If I were doing this I would use the methods described in the other answers.

Upvotes: 0

Jon Skeet
Jon Skeet

Reputation: 1500215

Convert the binary into a long value - you can use Long.parseLong(text, 2) for that if you've got the binary value as a string... and then Double.longBitsToDouble to perform the conversion to a double. For example:

public class Test {

  public static void main(String[] args) {
      String text = "0100000000001001001000011111101101010100010001000010110100011000";
      long parsedAsLong = Long.parseLong(text, 2);
      double converted = Double.longBitsToDouble(parsedAsLong);
      System.out.println(converted);
  }

}

Upvotes: 1

Louis Wasserman
Louis Wasserman

Reputation: 198033

Double.longBitsToDouble(long) takes a 64-bit long and converts it to a floating point double, which you can then print.

Given that, you can just use Long.parseLong(binary, 2) to convert a string of 0s and 1s to a long, and then use that method to get a floating point value.

Upvotes: 1

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