(scheme1) has lambda but NOT define


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1 CS61A Lecture Colleen Lewis (calc) review (scheme1) has lambda but NOT define Remember calcapply? STk> (calcapply '+ '(1 2 )) 6 STk> (calcapply '* '(2 4 )) 24 STk> (calcapply '/ '(10 2)) STk> (calcapply ' '( 2 1)) (calc) readevalprint loop (define (calc) (display "calc: ") (flush) (print (calceval (read))) (calc)) STk> (calc) calceval calc: (+ (* 2 ) 1 (define (calceval exp) ((number? exp) exp) ((list? exp) (calcapply (car exp) (map calceval (cdr exp)))) (else (error Calc: bad exp )))) 2 * + 4 (scheme1) DOES NOT HAVE DEFINE! STk> (scheme1) Scheme1: Scheme1: ( ) 1
2 Working with large programs! Start with small functions Understanding code Read it Recursively figure out any functions it calls Try to call the function to see what it does in different cases Trace the function and try to call it from the functions that call it. Approximate hierarchy of calls scheme1 apply1 substitute lookup maybequote lambdaexp? ifexp? quoteexp? expchecker constant? number? boolean? string? proceedure? lambdaexp? STk> (lambdaexp? ' (+ x 2))) #t STk> (lambdaexp? '(+ ) STk> (lambdaexp? '+) STk> (lambdaexp? '(lambda)) #t It isn t as picky as we might hope Write lambdaexp? in terms of expchecker (define (expchecker type) (lambda (exp) (and (pair? exp) (eq? (car exp) type)))) (lambdaexp? ' (+ x 2))) Did you write it with syntactic sugar? A) Yes B)No Some Helpers (define quoteexp? (expchecker 'quote)) (define ifexp? (expchecker 'if)) (define (constant? exp) (or (number? exp) (boolean? exp) (string? exp) (procedure? exp))) What is string? STk> (string? "hello") #t STk> (string? 12) STk> (string? 'hello) 2
3 (lookup name params args) STk> (lookup 'x STk> (lookup 'y '(x y) '(2 )) STk> (lookup 'y y STk> (lookup '* * Just returns it if it isn t in there lookup full functionality STk> (lookup 'fn This already works '(x fn) '( (lambda (y) (* y y)))) (lambda (y) (* y y)) STk> (lookup 'x '(cat)) (quote cat) cat was already a word, but we want to tell other people this thing IS ACTUALLY a word Full lookup (define (lookup name params args) ((null? params) name) ((eq? name (car params)) (maybequote (car args)) (else (lookup name (cdr params) (cdr args))))) maybequote (define (maybequote value) ((lambdaexp? value) value) ((constant? value) value) ((procedure? value) value) (else (list 'quote value)))) Substitution using substitute STk> ( ) '(* x x) '() (* ) STk> ((lambda (x y) (+ x y)) STk>(substitute '(+ x y) '(x y) '( (+ (lambda (y) (* x y)) ) ) What does this return? '() (lambda (y) (* y)) '(lambda (y) (* x y))
4 Scheme substitution review (* x x)) ) #[closure arglist=(x) ff1a1f8] STk> (( (* x x))) ) What does this return? A) B) 16 C)?? (* x x)) ) #[closure arglist=(x) ff1a1f8] ' '( (apply1 proc args) STk> (apply1 + '( ) STk> (apply1 Unlike calcapply apply1 can be called with REAL scheme functions ' Or lists representing functions Remember lambdaexp?? apply1 (define (apply1 proc args) ((procedure? proc) (apply proc args)) ((lambdaexp? proc) ( (substitute (caddr proc) (cadr proc) args )) (else (error "bad proc:" proc)))) scheme1 (define (scheme1) (display "Scheme1: ") (flush) (print ( (read))) (scheme1)) STk> ( ) ((constant? exp) exp) STk> ( '+) #[closure arglist=args ffde8] ((symbol? exp) (eval exp)) 4
5 STk> ( '(if (> )) ((ifexp? exp) (if ( (cadr exp)) ( (caddr exp)) ( (cadddr exp)))) STk> ( 'x) *** Error: unbound variable: x Current eval stack: 0 x 1 (eval exp) Things like + are quoted: '+ when they are passed to so this assumes x will be a variable not a word. with words STk> ( '(quote x)) x STk> ( ') STk> ( (quote (quote x))) x STk> ( ''x) x ((quoteexp? exp)(cadr exp)) These are all equivalent! ((lambdaexp? exp) exp) (define ( exp) ((constant? exp) exp) ((symbol? exp) (eval exp)) ((quoteexp? exp) (cadr exp)) ((ifexp? exp) (if ( (cadr exp)) ( (caddr exp)) ( (cadddr exp)))) ((lambdaexp? exp) exp) ((pair? exp) (else (error "?!?" exp)))) ((pair? exp) ) STk> ( '(+ 2 )) STk> ( '(+ ( 1) )) STk> ( '( ))
6 ((pair? exp) ) ((pair? exp) (apply1 ( (car exp)) (map (cdr exp)))) STk> ( '(+ 2 )) STk> ( '(+ ( 1) )) STk> ( '( )) lambdaexp? Solution (define (expchecker type) (lambda (exp) (and (pair? exp) (eq? (car exp) type)))) (define (lambdaexp? exp) ((expchecker 'lambda) exp)) (define lambdaexp? (expchecker 'lambda)) Write lookup (some functionality missing) (define (lookup name params args) ((null? params) name) ((eq? name (car params)) (car args)) (else (lookup name (cdr params) (cdr args))))) Substitution using substitute STk> ( ) '(* x x) '() (* ) STk> ((lambda (x y) (+ x y)) STk>(substitute '(+ x y) '(x y) '( (+ (lambda (y) (* x y)) ) ) What does this return? #[closure arglist=(y) ff1cc48] A procedure that takes argument y and adds to it '(lambda (y) (* x y)) '() (lambda (y) (* y)) (* x x)) ) #[closure arglist=(x) ff1a1f8] A recursive call will be made where bound will be ' '( 6
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