Вопрос по tiger-book Аппеля (8 глава, стр. 177).
(
cannon.ml:
http://ix.io/28an,
ir.ml:
http://ix.io/28ao)
Разбираюсь с реализацией ф-ции
reorder, вот её описание (я немного перефразировал для себя, но смысл не затронут):
Takes a [Ir.expr list] and returns a [(Ir.stmt, Ir.expr list)]. The first element [Ir.stmt] contains all the things that must be executed before the expression list [Ir.expr list]. This includes all the statement-parts of the [Ir.ESeq]'s, as well as any expressions to their left with which they did not commute.
(Remember, if [Ir.stmt] and [Ir.expr] don't "commute" then we can't change the order of their evaluation)
И пример:
[e1; e2; ESeq(s, e3)]
[s] must be pulled leftward past [e2] and [e1].
(0) If they all commute our [reorder] function will return
[(stmt, exprs)] where:
stmt := s
exprs := [e1; e2; e3]
(1) If [commute s e1 = true] and
[commute s e2 = false]
stmt := Seq(Move(t1, e1), Seq(Move(t2, e2), s))
exprs := [Temp t1; Temp t2; e3]
(2) If [commute s e1 = false] and
[commute s e2 = true]
stmt := Seq(Move(t1, e1), s)
exprs := [Temp t1; e2; e3]
(3) If all don't commute - same as (1)
Я не понимаю
(1), а именно почему нужно
Move(t1, e1).
Т.е. почему бы не переписать так
stmt := Seq(Move(t2, e2), s)
exprs := [e1; Temp t2; e3]
?
