The AddPermanent command creates a permanent entry associating key with value in a cache table.  The cache table can be given directly as cache, or cache can refer to a procedure that has, or can have, a cache remember table.  If such a procedure is given and it has a cache remember table, the permanent entry is inserted into that table. Otherwise, a new cache table is created, installed as the procedure's remember table and the permanent entry is inserted into that table.

To handle an expression sequence as a key, key must be given in a list. The actual key used for the entry is op(key).

If more than one argument is passed after key, then an expression sequence containing those arguments is used as value.

Assigning directly into a cache table, using the cache[key] := value notation, creates a temporary entry.  Calling AddPermanent is the only way to create a permanent entry into a cache table.

To remove a permanent entry from a cache table use RemovePermanent.

Creating a permanent entry for a key that already has a temporary entry is not an error.  In this case, the temporary entry is used.

For more information on cache tables, see Cache Package.

$\mathrm{c1}≔\mathrm{Cache}\left(\right)$

$\mathrm{c1}≔\mathrm{Cache}\left(512\right)$

$\mathrm{c1}\left[a\right]≔b$

${\mathrm{c1}}_a≔b$

$\mathrm{Cache}:-\mathrm{AddPermanent}\left(\mathrm{c1}\,\left[x\right]\,y\right)$

$\mathrm{Cache}:-\mathrm{AddPermanent}\left(\mathrm{c1}\,\left[y\right]\,z\right)$

$\mathrm{eval}\left(\mathrm{c1}\right)$

$\mathrm{Cache}\left(512\,'\mathrm{temporary}'=\left[a=b\right]\,'\mathrm{permanent}'=\left[y=z\,x=y\right]\right)$

p := proc(x,y) option cache; x+y; end proc;

$p≔\mathbf{proc}\left(x\,y\right)\mathbf{option}\mathrm{cache}\;x\+y\mathbf{end\; proc}$

$p\left(1\,1\right)$

$2$

$\mathrm{Cache}:-\mathrm{AddPermanent}\left(p\,\left[2\,2\right]\,8\right)$

$\mathrm{Cache}:-\mathrm{AddPermanent}\left(p\,\left[1\,2\right]\,5\right)$

$p\left(2\,2\right)$

$8$

$p\left(1\,2\right)$

$5$

$\mathrm{Cache}:-\mathrm{AddPermanent}\left(p\,\left[3\,3\right]\,5\,6\,7\right)$

$p\left(2\,2\right)$

$8$

$\mathrm{op}\left(4\,\mathrm{eval}\left(p\right)\right)$

$\mathrm{Cache}\left(512\,'\mathrm{temporary}'=\left[\left(1\,1\right)=2\right]\,'\mathrm{permanent}'=\left[\left(2\,2\right)=8\,\left(3\,3\right)=\left(5\,6\,7\right)\,\left(1\,2\right)=5\right]\right)$