Discussion 11: Macros
Switch to Pensieve:
- Everyone: Go to pensieve.co, log in with your @berkeley.edu email, and enter your group number (which was in the email that assigned you to this lab).
Once you're on Pensieve, you don't need to return to this page; Pensieve has all the same content (but more features). If for some reason Penseive doesn't work, return to this page and continue with the discussion.
Getting Started
To get help from a TA, If you do not have an in-person TA, you can reach your TA using this Zoom link.
If there are fewer than 3 people in your group, feel free to merge your group with another group in the room.
Everybody say your name, and then your favorite kind of pie.
Macros
A macro is a code transformation that is created using define-macro
and applied using a call expression. A macro call is evaluated by:
- Binding the formal paramters of the macro to the unevaluated operand expressions of the macro call.
- Evaluating the body of the macro, which returns an expression.
- Evaluating the expression returned by the macro in the environment of the original macro call.
scm> (define-macro (twice expr) (list 'begin expr expr))
twice
scm> (twice (+ 2 2)) ; evaluates (begin (+ 2 2) (+ 2 2))
4
scm> (twice (print (+ 2 2))) ; evaluates (begin (print (+ 2 2)) (print (+ 2 2)))
4
4
Debugging tip: In order to see what expression a macro creates, change it to a regular procedure, then call it with quoted arguments.
scm> (define (twice expr) (list 'begin expr expr)) ; Same definition, but with define instead of define-macro
twice
scm> (twice '(print (+ 2 2))) ; Called with a quoted argument
(begin (print (+ 2 2)) (print (+ 2 2)))
scm> (eval (twice '(print (+ 2 2)))) ; Evaluating the result has the same behavior as the original macro
4
4
scm> (define x (+ 2 1))
x
scm> `(x ,x)
(x 3)
scm> (define s '(1 2 3))
s
scm> `(+ x ,(cons '* s))
(+ x (* 1 2 3))
Q1: Mystery Macro
Figure out what this mystery-macro
does. Try to describe what
it does by reading the code and discussing examples as a group.
(define-macro (mystery-macro expr old new)
(mystery-helper expr old new))
(define (mystery-helper e o n)
(if (pair? e)
(cons (mystery-helper (car e) o n) (mystery-helper (cdr e) o n))
(if (eq? e o) n e)))
Pro Tip: Please don't just look at the hints right away. Hints are for when you get stuck.
mystery-macro
that could help you understand what it does and how it might be used.
scm> (define five 5)
five
scm> (mystery-macro (* x x) x five)
25
scm> (mystery-macro (* x x) x (+ five 1))
36
scm> (mystery-macro '(* x x) x y)
(* y y)
scm> (mystery-macro (> (x) (> (y) (+ x y))) > lambda)
(lambda (x) (lambda (y) (+ x y)))
scm> (mystery-macro (begin e e e) e (print five))
5
5
5
mystery-macro
replaces all instances of an old
symbol with a new
expression before evaluating the expression expr
.
Q2: Multiple Assignment
In Scheme, the expression returned by a macro procedure is evaluated in the
same environment in which the macro was called. Therefore, it's possible to return
a define
expression from a macro and have it affect the environment in which
the macro was called. This differs from a regular scheme procedure that contains
a define
expression, which would only affect the procedure's local frame.
In Python, we can bind two names to values in one line as follows:
>>> x, y = 1 + 1, 3 # now x is bound to 2 and y is bound to 3
>>> x, y = y, x # swap the values of x and y
>>> x
3
>>> y
2
Implement the assign
Scheme macro, which takes in two symbols sym1
and sym2
as well as two expressions expr1
and expr2
. It should bind
sym1
to the value of expr1
and sym2
to the value of expr2
in the
environment from which the macro was called.
scm> (assign x y (+ 1 1) 3) ; now x is bound to 2 and y is bound to 3
scm> (assign x y y x) ; swap the values of x and y
scm> x
3
scm> y
2
Make sure that expr2
is evaluated before sym1
is changed. Assume that
expr1
and expr2
do not have side effects (and so do not contain define
or
assign
expressions).
Call eval
on expr2
so that its value is included in the define
expression
created by assign
: ,(eval expr2)
. That way, the define
for expr1
won't
affect the value of expr2
, because expr2
will already have been evaluated.
Presentation Time: Come up with a one-sentence explanation of why the second define
line has to be different from the first define
line in this implementation. Choose someone from your group who hasn't presented recently to say this explanation to your TA for feedback in person or on Zoom.
For an optional extra challenge, try these additional tests that make
sure assign
works correctly even when the value of expr2
is not a number,
but instead a symbol.
(define z 'x) ; z is bound to the symbol x
(assign v w 2 z) ; now v is bound to 2 and w is bound to the symbol x
(assign v w w v) ; swap the values of v and w
(expect v x)
(expect w 2)
In order to ensure that the value of expr2
is not evaluated a second time, quote the result of evaluating it.
For example, (assign v w 2 z)
should be equivalent to:
(begin
(define v 2)
(define w (quote x)))
In this begin
expression, (quote x)
comes from first evaluating z
and then
quoting the result.
Q3: Switch
Define the macro switch
, which takes in an expression expr
and a list of
pairs called cases
where the first element of the pair is some number and the
second element is a single expression. switch
will evaluate the expression
contained in of cases
that corresponds to the number that expr
evaluates to.
scm> (switch (+ 1 1) ((1 (print 'a))
(2 (print 'b))
(3 (print 'c))))
b
You may assume that the value expr
evaluates to is always the first element of one of the pairs in cases
. You can also assume that the first value of each pair in
cases
is a number and the second expression does not contain the symbol val
.
Use equal?
to check if two numbers are equal.
(let ((val (+ 1 1)))
(cond ((equal? val 1) (print 'a))
((equal? val 2) (print 'b))
((equal? val 3) (print 'c))))
This expression first assigns val
to 3 and then compares val
to the first element in each pair in cases
.
Document the Occasion
Please all fill out the attendance form (one submission per person per week).