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double ScalarsScalars of type double are stored using eight bytes (64 bits). The format used corresponds to that of the IEEE754 standard and is stored little endian or LSB first. A double precision floatingpoint number is expressed as the product of two parts: the mantissa and a power of two. For example: ±mantissa × 2^{exponent} The mantissa represents the actual binary digits of the floatingpoint number. The power of two is represented by the exponent. The stored form of the exponent is an 11bit value from 0 to 2047. The actual value of the exponent is calculated by subtracting 1023 from the stored value (0 to 2047) giving a range of –1023 to +1024. The mantissa is a 53bit value (representing about thirteen decimal digits) whose most significant bit (MSB) is always 1 and is, therefore, not stored. There is also a sign bit that indicates whether the floatingpoint number is positive or negative. Double precision floatingpoint numbers are stored on word boundaries in the following format:
Where
Zero is a special value denoted with an exponent field of 0 and a mantissa of 0. Using the above format, the floatingpoint number 12.5 is stored as a hexadecimal value of 0xC029000000000000. In memory, this value appears as follows:
It is fairly simple to convert floatingpoint numbers to and from their hexadecimal storage equivalents. The following example demonstrates how this is done for the value 12.5 shown above. The floatingpoint storage representation is not an intuitive format. To convert this to a floatingpoint number, the bits must be separated as specified in the floatingpoint number storage format table shown above. For example:
From this illustration, you can determine the following:
There is an understood binary point at the left of the mantissa that is always preceded by a 1. This digit is omitted from the stored form of the floatingpoint number. Adding 1 and the binary point to the beginning of the mantissa gives the following value: 1.1001000000000000000000000000000000000000000000000000 To adjust the mantissa for the exponent, move the decimal point to the left for negative exponent values or right for positive exponent values. Since the exponent is three, the mantissa is adjusted as follows: 1100.1000000000000000000000000000000000000000000000000 The result is a binary floatingpoint number. Binary digits to the left of the decimal point represent the power of two corresponding to their position. For example, 1100 represents (1 × 2^{3}) + (1 × 2^{2}) + (0 × 2^{1}) + (0 × 2^{0}), which is 12. Binary digits to the right of the decimal point also represent the power of two corresponding to their position. However, the powers are negative. For example, .100... represents (1 × 2^{1}) + (0 × 2^{2}) + (0 × 2^{3}) + ... which equals .5. The sum of these values is 12.5. Because the sign bit was set, this number should be negative. So, the hexadecimal value 0xC029000000000000 is 12.5.  

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