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I have a string of arbitrary length representing a decimal integer value and converting this string into a big integer number in plain binary format (not BCD, more than 64 bits).
I am looking for a good simple estimate how many bytes will hold N decimal digits for sure without using floating-point arithmetics.
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The maximum number of bytes the database server uses to store a decimal value is 17. One byte is used to store the exponent and sign, leaving 16 bytes to store up to 32 digits of precision. If you specify a precision of 32 and an odd scale, however, you lose 1 digit of precision.
It takes on average 3.2 bits to represent a single decimal digit - 0 through 7 can be represented in 3 bits, while 8 and 9 require 4.
A byte is not just 8 values between 0 and 1, but 256 (28) different combinations (rather permutations) ranging from 00000000 via e.g. 01010101 to 11111111 . Thus, one byte can represent a decimal number between 0(00) and 255.
The most common grouping is 8 bits, which forms a byte. A single byte can represent 256 (28) numbers. Memory capacity is usually referred to in bytes.
Without using floating point arithmetic: For N decimal digits, you need
(98981 * N) / 238370 + 1
bytes. 98981/238370 is a good rational approximation (from above) to log(10)/log(256) (the 9th convergent), integer division truncates, therefore add 1. The formula is tight for N < 238370, the number of bytes needed to represent 10^N - 1 is exactly given by that, it overestimates for N a multiple of 238370 and really large N. If you're not too afraid of allocating the odd byte too much, you can also use (267 * N) / 643 + 1, (49 * N) / 118 + 1, (5 * N) / 12 + 1 or, wasting about 10% of space, (N + 1) / 2.
As @Henrick Hellström points out, in Delphi one would have to use the div operator for integer division (missed the delphi tag).
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