5.1 圖就不畫在機器上了，麻煩
5.2 用寄存器語言描述5.1題中的階乘機器，加上了讀取和打印,這里的解答全部在實際的寄存機器中驗證過，但是仍然按照該節(jié)的表示法表示。
(controller
fac-loop
(assign n (op read))
(assign product (const 1))
(assign counter (const 1))
iter-loop
(test (op >) (reg counter) (reg n))
(branch (label iter-done))
(assign product (op *) (reg product) (reg counter))
(assign counter (op +) (reg counter) (const 1))
(goto (label iter-loop))
iter-done
(perform (op print) (reg product))
(goto (label fac-loop)))
5.3 牛頓法求平方根，將這個過程轉(zhuǎn)化為寄存器語言，第一個版本，假設(shè)good-enough?和improve都是基本過程，
;version1
(controller
sqrt-loop
(test (op good-enough?) (reg guess))
(branch (label sqrt-done))
(assign guess (op improve) (reg guess))
(goto (label good-enough))
sqrt-done)
第二個版本，展開good-enough?過程，
;version2
(controller
good-enough
(assign t (op square) (reg guess))
(assign t (op -) (reg t) (reg x))
(assign t (op abs) (reg t))
(test (op <) (reg t) (const 0.001))
(branch (label sqrt-done))
(assign guess (op improve) (reg guess))
(goto (label good-enough))
sqrt-done)
最后，展開improve過程，
;version3
(controller
sqrt-init
(assign guess (const 1.0))
(assign x (op read))
good-enough
;good-enough
(assign t (op square) (reg guess))
(assign t (op -) (reg t) (reg x))
(assign t (op abs) (reg t))
(test (op <) (reg t) (const 0.001))
(branch (label sqrt-done))
;improve
(assign t (op /) (reg x) (reg guess))
(assign t (op +) (reg guess) (reg t))
(assign guess (op /) (reg t) (const 2.0))
(goto (label good-enough))
sqrt-done)
在start之后，從寄存器guess中得到最后結(jié)果。
5.4
a)第一個是一個指數(shù)計算過程，用到了遞歸，因此需要引入continue寄存器來保存和恢復(fù)堆棧，實現(xiàn)與階乘相似，如下
(controller
(assign continue (label expt-done))
expt-loop
(test (op =) (reg n) (const 0))
(branch (label expt-base-case))
;;保存continue
(save continue)
(assign n (op -) (reg n) (const 1))
(assign continue (label after-expt))
(goto (label expt-loop))
after-expt
;;恢復(fù)continue
(restore continue)
(assign val (op *) (reg b) (reg val))
(goto (reg continue))
expt-base-case
(assign val (const 1))
(goto (reg continue))
expt-done
(perform (op display) (reg val)))
b) 迭代型的遞歸計算過程，尾遞歸調(diào)用，因此不需要continue寄存器來保存和恢復(fù)堆棧，這道習(xí)題就是希望能分辨非尾遞歸和尾遞歸帶來的寄存機器語言的區(qū)別
(controller
(assign product (const 1))
(assign n (op read))
(assign b (op read))
(assign counter (reg n))
expt-iter-loop
(test (op =) (reg counter) (const 0))
(branch (label expt-done))
(assign counter (op -) (reg counter) (const 1))
(assign product (op *) (reg b) (reg product))
(goto (label expt-iter-loop))
expt-done
(perform (op display) (reg product)))
5.5 手工模擬就算了，5.2節(jié)就可以機器模擬了
5.6 是有一個多余的save和一個多余的restore操作，請看注釋：
(
(assign continue (label fib-done))
fib-loop
(test (op <) (reg n) (const 2))
(branch (label immediate-answer))
;;compute fib(n-1)
(save continue)
(assign continue (label after-fib-1))
(save n)
(assign n (op -) (reg n) (const 1))
(goto (label fib-loop))
after-fib-1
;;compute fib(n-2)
(restore n)
;這里多余，(restore continue)
(assign n (op -) (reg n) (const 2))
;這里多余，(save continue)
(assign continue (label after-fib-2))
(save val) ;;save fib(n-1)
(goto (label fib-loop))
after-fib-2
(assign n (reg val))
(restore val)
(restore continue)
(assign val (op +) (reg val) (reg n))
(goto (reg continue))
immediate-answer
(assign val (reg n))
(goto (reg continue))
fib-done)