Using Let in Scheme

问题:

I want to write a program to find the roots of the quadratic equation in Scheme. I used LET for certain bindings.

(define roots-with-let
  (λ (a b c)
    (let ((4ac (* 4 a c))
          (2a (* 2 a))
          (discriminant (sqrt ( - (* b b) (4ac)))))
      (cons ( / ( + (- b) discriminant) 2a)
            ( / ( - (- b) discriminant) 2a)))))

I defined the discriminant with 4ac since I did not want (* 4 a c). Even though I have defined (4ac (* 4 a c)), it is giving me this error:

expand: unbound identifier in module in: 4ac.

My question is how is let evaluated (what order)? And if i want 4ac in my let should i write another inner let? Is there a better way to do this?

回答1:

Use let* instead of let.

The difference between let and let* is the following:

let* binds variables from left to right. Earlier bindings can be used in new binding further to the right (or down).

let on the other hand can be thought of as syntactic sugar (or macro) for simple lambda abstraction:

(let ((a exp1)
      (b exp2))
   exp)

is equivalent to

((lambda (a b)
    exp)
 exp1 exp2)

回答2:

• 4ac is a variable with a numeric value, so (4ac) is not meaningful.

• LET binds all variables, but the variables can't be used in the computations for the values.

This does not work:

(let ((a 1) (b 1) (c (* a b)))
   c)

Use:

(let ((a 1) (b 1))
  (let ((c (* a b)))
    c))

Above introduces A and B with the first LET. In the second LET both A and B now can be used to compute C.

Or:

(let* ((a 1) (b 1) (c (* a b)))
   c)

回答3:

You'll need a special let-construct (let*) here since the variables inside the let-definition refer to each other.

It's rather a problem of defining a scope than of evaluating an expression (In usual let-definitions, the order of evaluation doesn't matter since the values may not use each other)

回答4:

When you use let, the bindings are not visible in any of the bodies. Use let* instead and see the RNRS docs for details.