Conditional Expressions and Predicates

SICP > Building Abstractions with Procedures > The Elements of Programming > Conditional Expressions and Predicates

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The expressive power of the class of procedures that we can define at this point is very limited, because we have no way to make tests and to perform different operations depending on the result of a test. For instance, we cannot define a procedure that computes the absolute value of a number by testing whether the number is positive, negative, or zero and taking different actions in the different cases according to the rule

$\begin{displaymath}\vert x\vert = \left\{ \begin{array}{rl} x & \mbox{if $x>0$ ... ...box{if $x=0$ } \\ -x & \mbox{if $x<0$ } \end{array} \right. \end{displaymath}$

This construct is called a case analysis, and there is a special form in Lisp for notating such a case analysis. It is called cond (which stands for ``conditional''), and it is used as follows:
(define (abs x)
  (cond ((> x 0) x)
        ((= x 0) 0)
        ((< x 0) (- x))))
The general form of a conditional expression is
(cond (p₁ e₁)
      (p₂ e₂)
      
      (p_n e_n))
consisting of the symbol cond followed by parenthesized pairs of expressions (p e) called clauses. The first expression in each pair is a predicate--that is, an expression whose value is interpreted as either true or false.
Conditional expressions are evaluated as follows. The predicate p₁ is evaluated first. If its value is false, then p₂ is evaluated. If p₂'s value is also false, then p₃ is evaluated. This process continues until a predicate is found whose value is true, in which case the interpreter returns the value of the corresponding consequent expression e of the clause as the value of the conditional expression. If none of the p's is found to be true, the value of the cond is undefined.
The word predicate is used for procedures that return true or false, as well as for expressions that evaluate to true or false. The absolute-value procedure abs makes use of the primitive predicates >, <, and =. These take two numbers as arguments and test whether the first number is, respectively, greater than, less than, or equal to the second number, returning true or false accordingly.
Another way to write the absolute-value procedure is
(define (abs x)
  (cond ((< x 0) (- x))
        (else x)))
which could be expressed in English as ``If x is less than zero return -x; otherwise return x.'' Else is a special symbol that can be used in place of the p in the final clause of a cond. This causes the cond to return as its value the value of the corresponding e whenever all previous clauses have been bypassed. In fact, any expression that always evaluates to a true value could be used as the p here.
Here is yet another way to write the absolute-value procedure:
(define (abs x)
  (if (< x 0)
      (- x)
      x))
This uses the special form if, a restricted type of conditional that can be used when there are precisely two cases in the case analysis. The general form of an if expression is
(if predicate consequent alternative)
To evaluate an if expression, the interpreter starts by evaluating the predicate part of the expression. If the predicate evaluates to a true value, the interpreter then evaluates the consequent and returns its value. Otherwise it evaluates the alternative and returns its value.
In addition to primitive predicates such as <, =, and >, there are logical composition operations, which enable us to construct compound predicates. The three most frequently used are these:

(and e₁ ... e_n)
The interpreter evaluates the expressions e one at a time, in left-to-right order. If any e evaluates to false, the value of the and expression is false, and the rest of the e's are not evaluated. If all e's evaluate to true values, the value of the and expression is the value of the last one.
(or e₁ ... e_n)
The interpreter evaluates the expressions e one at a time, in left-to-right order. If any e evaluates to a true value, that value is returned as the value of the or expression, and the rest of the e's are not evaluated. If all e's evaluate to false, the value of the or expression is false.
(not e)
The value of a not expression is true when the expression e evaluates to false, and false otherwise.

Notice that and and or are special forms, not procedures, because the subexpressions are not necessarily all evaluated. Not is an ordinary procedure.
As an example of how these are used, the condition that a number x be in the range 5 < x < 10 may be expressed as
(and (> x 5) (< x 10))
As another example, we can define a predicate to test whether one number is greater than or equal to another as
(define (>= x y)
  (or (> x y) (= x y)))
or alternatively as
(define (>= x y)
  (not (< x y)))
Exercise. Below is a sequence of expressions. What is the result printed by the interpreter in response to each expression? Assume that the sequence is to be evaluated in the order in which it is presented.
10

(+ 5 3 4)

(- 9 1)

(/ 6 2)

(+ (* 2 4) (- 4 6))

(define a 3)

(define b (+ a 1))

(+ a b (* a b))

(= a b)

(if (and (> b a) (< b (* a b)))
    b
    a)

(cond ((= a 4) 6)
      ((= b 4) (+ 6 7 a))
      (else 25))

(+ 2 (if (> b a) b a))

(* (cond ((> a b) a)
         ((< a b) b)
         (else -1))
   (+ a 1))
Exercise. Translate the following expression into prefix form

$\begin{displaymath}\frac{5+4+\left(2-\left(3-(6+\frac{4}{5})\right)\right)}{3 (6-2) (2-7)} \end{displaymath}$

Exercise. Define a procedure that takes three numbers as arguments and returns the sum of the squares of the two larger numbers.
Exercise. Observe that our model of evaluation allows for combinations whose operators are compound expressions. Use this observation to describe the behavior of the following procedure:
(define (a-plus-abs-b a b)
  ((if (> b 0) + -) a b))
Exercise. Ben Bitdiddle has invented a test to determine whether the interpreter he is faced with is using applicative-order evaluation or normal-order evaluation. He defines the following two procedures:
(define (p) (p))


(define (test x y)
  (if (= x 0)
      0
      y))
Then he evaluates the expression
(test 0 (p))
What behavior will Ben observe with an interpreter that uses applicative-order evaluation? What behavior will he observe with an interpreter that uses normal-order evaluation? Explain your answer. (Assume that the evaluation rule for the special form if is the same whether the interpreter is using normal or applicative order: The predicate expression is evaluated first, and the result determines whether to evaluate the consequent or the alternative expression.)

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