PLQ #9

Question 1 @ 2024-11-12 18:44

What’s the purpose of a compiler — what does it do?

(Choose the best answer.)

A Translate code in some language to machine code.
B Translate code in a high level language to a lower level language.
C Translate code in some language to some different language so it runs faster.
D Translate code to some other code, possibly in a different language.
E If we want to be precise, is any kind of code processing.

Question 2 @ 2024-11-12 18:45

What’s wrong with the following compiler?

(define (compile expr)
(cases expr
[(Num n) (lambda ([env : ENV]) (RktV n))]
[(Id name) (lambda ([env : ENV]) (lookup name env))]
...
[(Call fun-expr arg-exprs)
(define cfun (compile fun-expr))
(define compiled-args (map compile arg-exprs))
(lambda ([env : ENV])
(define fval ...)
(define arg-vals ...)
(cases fval
[(FunV names body fun-env)
(define compiled-body (compile body))
(compiled-body (extend names arg-vals fun-env))]
...))]
...))

A There is no problem.
B The use of compile is wrong, and will lead to a type error.
C Using compile at runtime is wrong, we should use eval instead.
D The compile function should never call itself.
E The compile function should never emit code that calls itself.

Question 3 @ 2024-11-12 18:48

What will the following evaluate to in the lazy PL language?

(define (my-if cond then else) (if cond then else))
(my-if (list (/ 1 0)) 0 1)

0
1
E0 Division by zero error
Es Syntax error due to using cond as a keyword
L Infinite loop

Question 4 @ 2024-11-12 18:49

We know that we can define if as a plain function in lazy Racket:

(define (my-if cond then else) (if cond then else))

And in fact, if (the builtin value) is itself a function. Can the same be said on cond?

Yes, No

Question 5 @ 2024-11-12 18:51

Assume this code in the lazy PL language:

(define foo (cons 1 (cons 2 (cons 3 bar))))
(define bar (cons 'boom (map add1 foo)))

How does the value that foo is bound to look like?

A It’s an infinite loop, nothing to see.
B Error due to an infinite loop
C Error due to trying to use add1 with a symbol
D (1 2 boom 3 4 boom 5 6 boom ...)
E (1 2 boom 2 3 boom 3 4 boom ...)
F (1 2 #<error> 3 4 #<error> 5 6 #<error> ...)
G (1 2 #<error> 2 3 #<error> 3 4 #<error> ...)
H (1 2 boom 3 4 #<error> 5 6 #<error> ...)
I (1 2 boom 2 3 #<error> 3 4 #<error> ...)

Question 6 @ 2024-11-12 18:54

Of the following Racket (or #lang pl) features, which ones are side-effects?