SICP Exercise 3.1 make-accumulator

Exercise 3.1.  An accumulator is a procedure that is called repeatedly with a single numeric argument and accumulates its arguments into a sum. Each time it is called, it returns the currently accumulated sum. Write a procedure make-accumulator that generates accumulators, each maintaining an independent sum. The input to make-accumulator should specify the initial value of the sum; for example

(define A (make-accumulator 5))
(A 10)
15
(A 10)
25

 SOLUTION

 The code is here:

Comments

Post a Comment

Popular posts from this blog

SICP Exercise 4.18 a alternative strategy for interpreting internal definitions

Exercise 4.18.   Consider an alternative strategy for scanning out definitions that translates the example in the text to (lambda < vars >   (let ((u '*unassigned*)         (v '*unassigned*))     (let ((a < e1 >)           (b < e2 >))       (set! u a)       (set! v b))     < e3 >)) Here  a  and  b  are meant to represent new variable names, created by the interpreter, that do not appear in the user's program. Consider the  solve  procedure from section  3.5.4 : (define (solve f y0 dt)   (define y (integral (delay dy) y0 dt))   (define dy (stream-map f y))   y) Will this ...
Read more

SICP Exercise 3.11 make-account internal definitions with local state

Image
Exercise 3.11.    In section  3.2.3  we saw how the environment model described the behavior of procedures with local state. Now we have seen how internal definitions work. A typical message-passing procedure contains both of these aspects. Consider the bank account procedure of section  3.1.1 : (define (make-account balance)   (define (withdraw amount)     (if (>= balance amount)         (begin (set! balance (- balance amount))                balance)         "Insufficient funds"))   (define (deposit amount)     (set! balance (+ balance amount))     balance)   (define (dispatch m)     (cond...
Read more

SICP Exercise 3.13 make-cycle

Image
Exercise 3.13.    Consider the following  make-cycle  procedure, which uses the  last-pair  procedure defined in exercise  3.12 : (define (make-cycle x)   (set-cdr! (last-pair x) x)   x) Draw a box-and-pointer diagram that shows the structure  z  created by (define z (make-cycle (list 'a 'b 'c))) What happens if we try to compute  (last-pair z) ? SOLUTION The code is here: Exercise 3.13 make-cycle The box-and-pointer diagrams are here:
Read more