4.9 Assignment: set!

Assignment: set! and set!-values in Reference: Racket also documents set!.

Assign to a variable using set!:

(set! id expr)

A set! expression evaluates expr and changes id (which must be bound in the enclosing environment) to the resulting value. The result of the set! expression itself is #<void>.

Examples:

(define greeted null)

(define (greet name)     (set! greeted (cons name greeted))     (string-append "Hello, " name))

> (greet "Athos")

"Hello, Athos"

> (greet "Porthos")

"Hello, Porthos"

> (greet "Aramis")

"Hello, Aramis"

> greeted

'("Aramis" "Porthos" "Athos")

(define (make-running-total)     (let ([n 0])       (lambda ()         (set! n (+ n 1))         n)))   (define win (make-running-total))   (define lose (make-running-total))

> (win)   1   > (win)   2   > (lose)   1   > (win)   3

4.9.1 Guidelines for Using Assignment

Although using set! is sometimes appropriate, Racket style generally discourages the use of set!. The following guidelines may help explain when using set! is appropriate.

• As in any modern language, assigning to shared identifier is no substitute for passing an argument to a procedure or getting its result.

  Really awful example:

  (define name "unknown")   (define result "unknown")   (define (greet)     (set! result (string-append "Hello, " name)))     > (set! name "John")   > (greet)   > result   "Hello, John"

  Ok example:

  (define (greet name)     (string-append "Hello, " name))     > (greet "John")   "Hello, John"   > (greet "Anna")   "Hello, Anna"

• A sequence of assignments to a local variable is far inferior to nested bindings.

  Bad example:

  > (let ([tree 0])       (set! tree (list tree 1 tree))       (set! tree (list tree 2 tree))       (set! tree (list tree 3 tree))       tree)   '(((0 1 0) 2 (0 1 0)) 3 ((0 1 0) 2 (0 1 0)))

  Ok example:

  > (let* ([tree 0]            [tree (list tree 1 tree)]            [tree (list tree 2 tree)]            [tree (list tree 3 tree)])       tree)   '(((0 1 0) 2 (0 1 0)) 3 ((0 1 0) 2 (0 1 0)))

• Using assignment to accumulate results from an iteration is bad style. Accumulating through a loop argument is better.

  Somewhat bad example:

  (define (sum lst)     (let ([s 0])       (for-each (lambda (i) (set! s (+ i s)))                 lst)       s))     > (sum '(1 2 3))   6

  Ok example:

  (define (sum lst)     (let loop ([lst lst] [s 0])       (if (null? lst)           s           (loop (cdr lst) (+ s (car lst))))))     > (sum '(1 2 3))   6

  Better (use an existing function) example:

  (define (sum lst)     (apply + lst))     > (sum '(1 2 3))   6

  Good (a general approach) example:

  (define (sum lst)     (for/fold ([s 0])               ([i (in-list lst)])       (+ s i)))     > (sum '(1 2 3))   6

• For cases where stateful objects are necessary or appropriate, then implementing the object’s state with set! is fine.

  Ok example:

  (define next-number!     (let ([n 0])       (lambda ()         (set! n (add1 n))         n)))     > (next-number!)   1   > (next-number!)   2   > (next-number!)   3

All else being equal, a program that uses no assignments or mutation is always preferable to one that uses assignments or mutation. While side effects are to be avoided, however, they should be used if the resulting code is significantly more readable or if it implements a significantly better algorithm.

The use of mutable values, such as vectors and hash tables, raises fewer suspicions about the style of a program than using set! directly. Nevertheless, simply replacing set!s in a program with a vector-set!s obviously does not improve the style of the program.

4.9.2 Multiple Values: set!-values

Assignment: set! and set!-values in Reference: Racket also documents set!-values.

The set!-values form assigns to multiple variables at once, given an expression that produces an appropriate number of values:

(set!-values (id ...) expr)

This form is equivalent to using let-values to receive multiple results from expr, and then assigning the results individually to the ids using set!.

Examples:
(define game     (let ([w 0]           [l 0])       (lambda (win?)         (if win?             (set! w (+ w 1))             (set! l (+ l 1)))         (begin0           (values w l)           ; swap sides...           (set!-values (w l) (values l w))))))
> (game #t)
1
0
> (game #t)
1
1
> (game #f)
1
2