CODEWITHSUNDEEP
java
user3105533
user3105533

Reputation: 331

Cannot convert from a double number to binary

I have the next code which tries to convert a double type number into its binary form. I know the algorithm, the integer part of the number can be converted to binary the same way an int type number would, and the remainig fractionary part can be converted this way:

    integerPart = (int)value;
    fractionaryPart = value-integerPart;

    while(fractionaryPart != 0.0)
    {
        fractionaryPart = fractionaryPart*2;
        integerPart = (int)fractionaryPart;
        fractionaryPart = fractionaryPart - integerPart; 
    }

this is the code to convert the integer part

bits = 64 //8 bytes
char binary[] = new char[bits];
for(i = bits-1;i >= 0;i--)
{
    if((integerPart&1) == 1)
       binary[i] = '1';
    else if((integerPart&1) == 0)
       binary[i] = '0';
    integerPart >>= 1;  
}

the problem is that I have to print the entire number the way a double number would look into the RAM memory (an 8 bytes format) for example if I have the number 42.375, the result must be 101010.011. But I don´t know how to combine the binary numbers of the while() and the for() and put them into an array of 64 bits, like:

|00000000|00000000|00000000|00000000|00000000|00000000|00000010|1010.011|

any suggestions??

Upvotes: 2

Views: 273

Answers (2)

Louis Wasserman
Louis Wasserman

Reputation: 198143

That is not how doubles are represented in memory.

float representation

The above representation is for a 32-bit float, for a double it is 52 bits of mantissa, 11 bits of exponent, and 1 sign bit.

Double.doubleToLongBits is how you get this out, and longBitsToDouble is the reverse conversion. The Javadoc of longBitsToDouble discusses the details of the conversion.

Upvotes: 3

Atsby
Atsby

Reputation: 2327

The way I've always printed a double number the way it looks in RAM is using one of the Double.doubleToLongBits or Double.doubleToRawLongBits methods in the Double class.

Upvotes: 0

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