Re: reducing NUMERIC size for 9.1

On Fri, Jul 30, 2010 at 2:08 PM, Tom Lane <tgl(at)sss(dot)pgh(dot)pa(dot)us> wrote:
> Robert Haas <robertmhaas(at)gmail(dot)com> writes:
>> ....  Maybe something like this,
>> obviously with a suitable comment which I haven't written yet:
>
>>     numeric_digits = (precision + 6) / 4;
>>     return (numeric_digits * sizeof(int16)) + NUMERIC_HDRSZ;
>
> This is OK for the base-10K case, but there's still code in there
> for the base-10 and base-100 cases.  Can you express this logic in
> terms of DEC_DIGITS and sizeof(NumericDigit) ?  I think you might
> find it was actually clearer that way, cf Polya.

It appears to work out to:

numeric_digits = (precision + 2 * (DEC_DIGITS - 1)) / DEC_DIGITS
return (numeric_digits * sizeof(NumericDigits)) + NUMERIC_HDRSZ;

The smallest value for precision which requires 2 numeric_digits is
always 2; and the required number of numeric_digits increases by 1
each time the number of base-10 digits increases by DEC_DIGITS.

--
Robert Haas
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