Decimal64 floating-point format

In computing, decimal64 is a decimal floating-point computer numbering format that occupies 8 bytes (64 bits) in computer memory. It is intended for applications where it is necessary to emulate decimal rounding exactly, such as financial and tax computations.

Decimal64 supports 16 decimal digits of significand and an exponent range of −383 to +384, i.e. ±0.000  000  000  000  000 to ±9.999  999  999  999  999. (Equivalently, ±0 000  000  000  000  000 to ±9  999  999  999  999  999.) In contrast, the corresponding binary format, which is the most commonly used type, has an approximate range of ±0.000  000  000  000  001 to ±1.797  693  134  862  315. Because the significand is not normalized, most values with less than 16 significant digits have multiple possible representations; 1 × 102 =  0.1 × 103  =  0.01 × 104, etc. Zero has 768 possible representations (1536 if both signed zeros are included).

Decimal64 floating point is a relatively new decimal floating-point format, formally introduced in the 2008 version of IEEE 754 as well as with ISO/IEC/IEEE 60559:2011.

Representation of decimal64 values
IEEE 754 allows two alternative representation methods for decimal64 values. The standard does not specify how to signify which representation is used, for instance in a situation where decimal64 values are communicated between systems:


 * In the binary representation method, the 16-digit significand is represented as a binary coded positive integer, based on binary integer decimal (BID).
 * In the decimal representation method, the 16-digit significand is represented as a decimal coded positive integer, based on densely packed decimal (DPD) with 5 groups of 3 digits (except the most significant digit encoded specially) are each represented in declets (10-bit sequences). This is pretty efficient, because 210 = 1024, is only little more than needed to still contain all numbers from 0 to 999.

Both alternatives provide exactly the same range of representable numbers: 16 digits of significand and $3 × 2^{8} = 768$ possible decimal exponent values. (All the possible decimal exponent values storable in a binary64 number are representable in decimal64, and most bits of the significand of a binary64 are stored keeping roughly the same number of decimal digits in the significand.)

In both cases, the most significant 4 bits of the significand (which actually only have 10 possible values) are combined with the most significant 2 bits of the exponent (3 possible values) to use 30 of the 32 possible values of a 5-bit field. The remaining combinations encode infinities and NaNs.

In the cases of Infinity and NaN, all other bits of the encoding are ignored. Thus, it is possible to initialize an array to Infinities or NaNs by filling it with a single byte value.

Binary integer significand field
This format uses a binary significand from 0 to $10^{16} − 1 = 9 999 999  999  999  999 = 2386F26FC0FFFF_{16} = 1000  1110000110  1111001001  1011111100  0000111111  1111111111_{2}.$

The encoding, completely stored on 64 bits, can represent binary significands up to $10 × 2^{50} − 1 = 11 258 999  068  426  239 = 27FFFFFFFFFFFF_{16},$ but values larger than $10^{16} − 1$ are illegal (and the standard requires implementations to treat them as 0, if encountered on input).

As described above, the encoding varies depending on whether the most significant $4 bits$ of the significand are in the range 0 to 7 (00002 to 01112), or higher (10002 or 10012).

If the 2 after the sign bit are "00", "01", or "10", then the exponent field consists of the $10 bits$ following the sign bit, and the significand is the remaining $53 bits$, with an implicit leading $0 bit$:

s 00eeeeeeee  (0)ttt tttttttttt tttttttttt tttttttttt tttttttttt tttttttttt s 01eeeeeeee  (0)ttt tttttttttt tttttttttt tttttttttt tttttttttt tttttttttt s 10eeeeeeee  (0)ttt tttttttttt tttttttttt tttttttttt tttttttttt tttttttttt This includes subnormal numbers where the leading significand digit is 0.

If the $2 bits$ after the sign bit are "11", then the 10-bit exponent field is shifted $2 bits$ to the right (after both the sign bit and the "11" bits thereafter), and the represented significand is in the remaining $51 bits$. In this case there is an implicit (that is, not stored) leading 3-bit sequence "100" for the most bits of the true significand (in the remaining lower bits ttt...ttt of the significand, not all possible values are used).

s 1100eeeeeeee (100)t tttttttttt tttttttttt tttttttttt tttttttttt tttttttttt s 1101eeeeeeee (100)t tttttttttt tttttttttt tttttttttt tttttttttt tttttttttt s 1110eeeeeeee (100)t tttttttttt tttttttttt tttttttttt tttttttttt tttttttttt

The 2-bit sequence "11" after the sign bit indicates that there is an implicit 3-bit prefix "100" to the significand. Compare having an implicit 1-bit prefix "1" in the significand of normal values for the binary formats. The 2-bit sequences "00", "01", or "10" after the sign bit are part of the exponent field.

The leading bits of the significand field do not encode the most significant decimal digit; they are simply part of a larger pure-binary number. For example, a significand of 8 000  000  000  000  000 is encoded as binary 0111  0001101011  1111010100  1001100011  0100000000  00000000002, with the leading $4 bits$ encoding 7; the first significand which requires a 54th bit is $2^{53} = 9 007 199  254  740  992.$ The highest valid significant is 9  999  999  999  999  999 whose binary encoding is (100)0  1110000110  1111001001  1011111100  0000111111  11111111112 (with the 3 most significant bits (100) not stored but implicit as shown above; and the next bit is always zero in valid encodings).

In the above cases, the value represented is



If the four bits after the sign bit are "1111" then the value is an infinity or a NaN, as described above:

0 11110 xx...x   +infinity 1 11110 xx...x   -infinity x 11111 0x...x   a quiet NaN x 11111 1x...x   a signalling NaN

Densely packed decimal significand field
In this version, the significand is stored as a series of decimal digits. The leading digit is between 0 and 9 (3 or 4 binary bits), and the rest of the significand uses the densely packed decimal (DPD) encoding.

The leading $2 bits$ of the exponent and the leading digit (3 or $4 bits$) of the significand are combined into the five bits that follow the sign bit.

This eight bits after that are the exponent continuation field, providing the less-significant bits of the exponent.

The last $50 bits$ are the significand continuation field, consisting of five 10-bit declets. Each declet encodes three decimal digits using the DPD encoding.

If the first two bits after the sign bit are "00", "01", or "10", then those are the leading bits of the exponent, and the three bits "TTT" after that are interpreted as the leading decimal digit (0 to 7):

s 00 TTT (00)eeeeeeee (0TTT)[tttttttttt][tttttttttt][tttttttttt][tttttttttt][tttttttttt] s 01 TTT (01)eeeeeeee (0TTT)[tttttttttt][tttttttttt][tttttttttt][tttttttttt][tttttttttt] s 10 TTT (10)eeeeeeee (0TTT)[tttttttttt][tttttttttt][tttttttttt][tttttttttt][tttttttttt]

If the first two bits after the sign bit are "11", then the second 2-bits are the leading bits of the exponent, and the next bit "T" is prefixed with implicit bits "100" to form the leading decimal digit (8 or 9):

s 1100 T (00)eeeeeeee (100T)[tttttttttt][tttttttttt][tttttttttt][tttttttttt][tttttttttt] s 1101 T (01)eeeeeeee (100T)[tttttttttt][tttttttttt][tttttttttt][tttttttttt][tttttttttt] s 1110 T (10)eeeeeeee (100T)[tttttttttt][tttttttttt][tttttttttt][tttttttttt][tttttttttt]

The remaining two combinations (11 110 and 11 111) of the 5-bit field after the sign bit are used to represent ±infinity and NaNs, respectively.

The DPD/3BCD transcoding for the declets is given by the following table. b9...b0 are the bits of the DPD, and d2...d0 are the three BCD digits.

The 8 decimal values whose digits are all 8s or 9s have four codings each. The bits marked x in the table above are ignored on input, but will always be 0 in computed results. (The $(−1)^{sign} × 10^{exponent−398} × significand$ non-standard encodings fill in the gap between $8 × 3 = 24$)

In the above cases, with the true significand as the sequence of decimal digits decoded, the value represented is


 * $$(-1)^\text{signbit}\times 10^{\text{exponentbits}_2-398_{10}}\times \text{truesignificand}_{10}$$