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Kirill Smelkov
mariadb
Commits
10db96fe
Commit
10db96fe
authored
Feb 20, 2005
by
unknown
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decimal format documented
parent
1683fc96
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strings/decimal.c
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strings/decimal.c
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10db96fe
...
@@ -107,6 +107,21 @@
...
@@ -107,6 +107,21 @@
#include <m_string.h>
#include <m_string.h>
#include <decimal.h>
#include <decimal.h>
/*
Internally decimal numbers are stored base 10^9 (see DIG_BASE below)
So one "decimal_digit" is
0 < decimal_digit <= DIG_MAX < DIG_BASE
in the struct st_decimal:
intg is the number of *decimal* digits (NOT number of decimal_digit's !)
before the point
frac - number of decimal digits after the point
buf is an array of decimal_digit's
len is the length of buf (length of allocated space) in decimal_digit's,
not in bytes
*/
typedef
decimal_digit
dec1
;
typedef
decimal_digit
dec1
;
typedef
longlong
dec2
;
typedef
longlong
dec2
;
...
@@ -1073,6 +1088,68 @@ int decimal2longlong(decimal *from, longlong *to)
...
@@ -1073,6 +1088,68 @@ int decimal2longlong(decimal *from, longlong *to)
RETURN VALUE
RETURN VALUE
E_DEC_OK/E_DEC_TRUNCATED/E_DEC_OVERFLOW
E_DEC_OK/E_DEC_TRUNCATED/E_DEC_OVERFLOW
DESCRIPTION
for storage decimal numbers are converted to the "binary" format.
This format has the following properties:
1. length of the binary representation depends on the {precision, scale}
as provided by the caller and NOT on the intg/frac of the decimal to
convert.
2. binary representations of the same {precision, scale} can be compared
with memcmp - with the same result as decimal_cmp() of the original
decimals (not taking into account possible precision loss during
conversion).
This binary format is as follows:
1. First the number is converted to have a requested precision and scale.
2. Every full DIG_PER_DEC1 digits of intg part are stored in 4 bytes
as is
3. The first intg % DIG_PER_DEC1 digits are stored in the reduced
number of bytes (enough bytes to store this number of digits -
see dig2bytes)
4. same for frac - full decimal_digit's are stored as is,
the last frac % DIG_PER_DEC1 digits - in the reduced number of bytes.
5. If the number is negative - every byte is inversed.
5. The very first bit of the resulting byte array is inverted (because
memcmp compares unsigned bytes, see property 2 above)
Example:
1234567890.1234
internally is represented as 3 decimal_digit's
1 234567890 123400000
(assuming we want a binary representation with precision=14, scale=4)
in hex it's
00-00-00-01 0D-FB-38-D2 07-5A-EF-40
now, middle decimal_digit is full - it stores 9 decimal digits. It goes
into binary representation as is:
........... 0D-FB-38-D2 ............
First decimal_digit has only one decimal digit. We can store one digit in
one byte, no need to waste four:
01 0D-FB-38-D2 ............
now, last digit. It's 123400000. We can store 1234 in two bytes:
01 0D-FB-38-D2 04-D2
So, we've packed 12 bytes number in 7 bytes.
And now we invert the highest bit to get the final result:
81 0D FB 38 D2 04 D2
And for -1234567890.1234 it would be
7E F2 04 37 2D FB 2D
*/
*/
int
decimal2bin
(
decimal
*
from
,
char
*
to
,
int
precision
,
int
frac
)
int
decimal2bin
(
decimal
*
from
,
char
*
to
,
int
precision
,
int
frac
)
{
{
...
...
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