Thread: Explanation of pg_column_size

Explanation of pg_column_size

From
Steve Baldwin
Date:
Hi all,

Can someone please explain the following to me:

mse=# with l as (select 0.1::numeric as numlit) select pg_column_size(numlit) as sz_numlit, pg_column_size(0.1::numeric) as sz_expr from l;
 sz_numlit | sz_expr
-----------+---------
         5 |       8
(1 row)

It looks as though the expression is being cast to double precision, but maybe it is invalid to use pg_column_size in this way.

I'm using 9.5.3 on OS X.

Thanks,

Steve

Re: Explanation of pg_column_size

From
Tom Lane
Date:
Steve Baldwin <steve.baldwin@gmail.com> writes:
> Can someone please explain the following to me:

> mse=# with l as (select 0.1::numeric as numlit) select
> pg_column_size(numlit) as sz_numlit, pg_column_size(0.1::numeric) as
> sz_expr from l;
>  sz_numlit | sz_expr
> -----------+---------
>          5 |       8
> (1 row)

I think the first case is putting the value into a tuple, which will cause
it to be trimmed to short-header form (1 length byte, 4 payload bytes).
In the other case you're looking at a raw numeric_in result, which will
be in the default 4-byte-length-word format (plus the same 4 payload
bytes).  The former would be a more accurate representation of how
big this specific value would be on-disk.

            regards, tom lane