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Most numeric programs involve converting between base-2 floating-point numbers, as represented by the computer, and base-10 floating-point numbers, as entered and handled by the programmer. What might not be obvious is the number of base-2 bits vs. base-10 digits required for each representation. Consider the following tables showing the number of decimal digits representable in a given number of bits, and vice versa:

binary in | 24 | 53 | 64 | 113 | 237 |

decimal out | 9 | 17 | 21 | 36 | 73 |

decimal in | 7 | 16 | 34 | 70 |

binary out | 25 | 55 | 114 | 234 |

We can compute the table numbers with these two functions:

int matula(int nbits) { /* Return output decimal digits needed for nbits-bits input. */ return ((int)ceil((double)nbits / log2(10.0) + 1.0)); } int goldberg(int ndec) { /* Return output bits needed for ndec-digits input. */ return ((int)ceil((double)ndec / log10(2.0) + 1.0)); }

One significant observation from those numbers is that we cannot achieve correct round-trip conversion between the decimal and binary formats in the same storage size! For example, we need 25 bits to represent a 7-digit value from the 32-bit decimal format, but the binary format only has 24 available. Similar observations hold for each of the other conversion pairs.

The general input/output base-conversion problem is astonishingly complicated, and solutions were not generally known until the publication of two papers in 1990 that are listed later near the end of this chapter. For the 128-bit formats, the worst case needs more than 11,500 decimal digits of precision to guarantee correct rounding in a binary-to-decimal conversion!

For further details see the references for Bennett Goldberg and David Matula.

Next: Further Reading, Previous: Complex Arithmetic, Up: Floating Point in Depth [Contents][Index]