Question

Suppose I have a table with a column name varchar(20), and I store a row with name = “abcdef”.

INSERT INTO tab(id, name) values(12, 'abcdef');

How is the memory allocation for name done in this case?

There are two ways I can think of:

a)

20 bytes is allocated but only 6 used. In this case varchar2 does not have any significant advantage over char, in terms of memory allocation.

b)

Only 6 bytes is allocated. If this is the case, and I addded a couple of more rows after this one,

INSERT INTO tab(id, name) values(13, 'yyyy');
INSERT INTO tab(id, name) values(14, 'zzzz');

and then I do a UPDATE,

UPDATE tab SET name = 'abcdefghijkl' WHERE id = 12;

Where does the DBMS get the extra 6 bytes needed from? There can be a case that the next 6 bytes are not free (if only 6 were allocated initially, next bytes might have been allotted for something else).

Is there any other way than shifting the row out to a new place? Even shifting would be a problem in case of index organized tables (it might be okay for heap organized tables).

Answer 1

There may be variations depending on the rdbms you are using, but generally:

Only the actual data that you store in a varchar field is allocated. The size is only a maximum allowed, it’s not how much is allocated.

I think that goes for char fields also, on some systems. Variable size data types are handled efficiently enough that there is no longer any gain in allocating the maximum.

If you update a record so that it needs more space, the record inside the same allocation block are moved down, and if the records no longer fit in the block, another block is allocated and the records are distributed between the blocks. That means that records are continous inside the allocation blocks, but the blocks doesn’t have to be continous on the disk.