Consider the following extract from a program:

$1$ c l a s s Shape

$2\{$

$3$ public :

$4$ Shape $()$ ;

$5$ void s e t S i z e $($ int $)$ ;

$6$ v i rtua l void draw $()=0$ ;

$7$ protected :

$8$ int s i z e ;

$9\}$ ;

$10$ c l a s s Square : public Shape

$11\{$

$12$ public :

$13$ void draw $()$ ;

$14\}$ ;

$15$ c l a s s Line : public Shape

$16\{$

$17$ public :

$18$ bool draw $()$ ;

$19\}$ ;

$($a$)$ Waarom word draw$()$ in die Shape klas virtual verklaar? Wat beteken dit?

Why is draw$()$ in the Shape class declared virtual? What does it mean?

Skep vtable, maak polimorfisme moontlik, sodat basisklas pointer na afgeleide

klas objek afgeleide klas impl kan sien $(3)$

$($b$)$ Watter draw$()$ implementering sou geroep word as $$n Shape pointer na $$n Square objek

gebruik word om draw$()$ te roep? Motiveer.

Which draw$()$ implementation would be called when a Shape pointer to a Square object

was used to call draw$()?$ Motivate.

Afgeleide klas, virtual in basisklas en signature stem ooreen in afgeleide klas $(4)$

$($c$)$ Watter draw$()$ implementering sou geroep word as $$n Shape pointer na $$n Line objek

gebruik word om draw$()$ te roep? Motiveer.

Which draw$()$ implementation would be called when a Shape pointer to a Line object

was used to call draw$()?$ Motivate.

Geen: Basisklas is suiwer, geen impl, en afgeleide klas het nie $$n void draw

nie. Fout kom voor wanneer Line objek gemaak word.