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2.14 (a) LL: E -> (E)E | [E]E | {E}E | ɛ LR: E -> E(E) | E[E] | E{E} | ɛ

(b)

( [ { ) ] } ɛ

E (E)E [E]E {E}E - - - ɛ

( [ { ) ] } ɛ

E E(E) E[E] E{E} - - - ɛ

(c) LL Tree:

(d) E -> (E -> ([E -> ([]E -> ([](E -> ([]([E -> ([]([]E -> ([]([])E -> ([]([]))E -> ([]([]))[E-> ([]([]))[]E -> ([]([]))[](E -> ([]([]))[]((E -> ([]([]))[](()E -> ([]([]))[](())E -> ([]([]))[](())

LR Tree

(d) E -> E) -> E)) -> E()) -> E(()) -> E](()) -> E[](()) -> E)[](()) -> E))[](()) -> E]))[](()) -> E]))[](()) -> E[]))[](()) -> E([]))[](()) -> E]([]))[](()) -> E[]([]))[](()) -> E([]([]))[](()) -> ([]([]))[](())

3.7 (a) When he calls the reverse function, it creates a new list and the original unreversed list remains loaded in memory, causing a leak.

(b) The reverse function changes the pointer passed to it to point to the tail of the newly reversed list, so when he tries delete_list(L), it deletes the reversed list he just created. Then, after assigning L = T, L points to the deallocated tail of the list he just deleted.

3.14 Static: 1122 Dynamic: 1121, with a static scope, every time set_x(n) is called the global variable x is changed, so the final print_x() will reflect that the second() function changed x to 2. With the dynamic scope, after second() is called only the local x is changed to 2, so the final print_x() call prints the global variable x which is still 1.