Oracle NUMBER(p) storage size?

2019-03-18 12:18发布

I've searched for it but i can't find a conclusive answer to my question...

I need to know what is the storage size of a number(p) field in Oracle.

Examples: NUMBER(1), NUMBER(3), NUMBER(8), NUMBER(10) etc...

3条回答
ゆ 、 Hurt°
2楼-- · 2019-03-18 12:27

The number data type in Oracle is a special data type that is variable length like varchar. Then if you store the same data in number(5) and number(20) the storage is the same like declaring a column as varchar(100) and varchar(200).

So specifying the p parameter in number(p,s) has no effect on storage size and is only for applying constraint on the data. But specifying the s parameter can reduce the size by rounding the data.

the minimum storage size of number data type is 1 byte and the maximum is 21 bytes. So if you do not want to apply constraint then use number data type without p parameter.

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男人必须洒脱
3楼-- · 2019-03-18 12:28
NUMBER  
999...(38 9's) x10125 
maximum value   Can be represented to full 38-digit precision (the mantissa).

-999...(38 9's) x10125 
minimum value   Can be represented to full 38-digit precision (the mantissa).


Precision   38 significant digits    ==> NUMBER(38) is the max

Refer here and also may be here

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一纸荒年 Trace。
4楼-- · 2019-03-18 12:31

The storage used depends on the actual numeric value, as well as the column precision and scale of the column.

The Oracle 11gR2 concepts guide says:

Oracle Database stores numeric data in variable-length format. Each value is stored in scientific notation, with 1 byte used to store the exponent. The database uses up to 20 bytes to store the mantissa, which is the part of a floating-point number that contains its significant digits. Oracle Database does not store leading and trailing zeros.

The 10gR2 guide goes further:

Taking this into account, the column size in bytes for a particular numeric data value NUMBER(p), where p is the precision of a given value, can be calculated using the following formula:

ROUND((length(p)+s)/2))+1

where s equals zero if the number is positive, and s equals 1 if the number is negative.

Zero and positive and negative infinity (only generated on import from Version 5 Oracle databases) are stored using unique representations. Zero and negative infinity each require 1 byte; positive infinity requires 2 bytes.

If you have access to My Oracle Support, there is more information in note 1031902.6.

You can see the actual storage used with vsize or dump.

create table t42 (n number(10));

insert into t42 values (0);
insert into t42 values (1);
insert into t42 values (-1);
insert into t42 values (100);
insert into t42 values (999);
insert into t42 values (65535);
insert into t42 values (1234567890);

select n, vsize(n), dump(n)
from t42
order by n;

          N   VSIZE(N)                           DUMP(N) 
------------ ---------- ---------------------------------
         -1          3           Typ=2 Len=3: 62,100,102 
          0          1                  Typ=2 Len=1: 128 
          1          2                Typ=2 Len=2: 193,2 
        100          2                Typ=2 Len=2: 194,2 
        999          3           Typ=2 Len=3: 194,10,100 
      65535          4          Typ=2 Len=4: 195,7,56,36 
 1234567890          6   Typ=2 Len=6: 197,13,35,57,79,91 

Notice that the storage varies depending on the value, even though they are all in a number(10) column, and that two 3-digit numbers can need different amounts of storage.

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