Fundamentals of Computer Science I: Media Computing (CS151.02 2007F)
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This reading is also available in PDF.
Summary: Many programs need to make choices. In this reading, we consider Scheme's conditional expressions, expressions that allow programs to behave differently in different situations. We also consider how conditionals help us draw some simple shapes.
Contents:
cond
and and or
When Scheme encounters a procedure call, it looks at all of the subexpressions within the parentheses and evaluates each one. Sometimes, however, the programmer wants Scheme to exercise more discretion. Specifically, the programmer wants to select just one subexpression for evaluation from two or more alternatives. In such cases, one uses a conditional expression, an expression that tests whether some condition is met and selects the subexpression to evaluate on the basis of the outcome of that test.
For instance, suppose we want to increase a value by 20% if it is greater than 127 and reduce it by 20% if it is less than 128. (You may recall this problem from a recent laboratory exercise.) While it is possible to write an interesting expression to do this computation, many programmers would prefer something a bit clearer.
To write a procedure that like this, we benefit from a mechanism that allows us to explicitly tell Scheme how to choose which expression to evaluate. Such mechanisms are the primary subject of this reading.
The simplest conditional expression in Scheme is an if expression. An if expression typically has three components: a test, a consequent, and an alternative. It selects one or the other of these expressions, depending on the outcome of a test. The general form is
(if test consequent alternative)
We'll return to the particular details in a moment. For now, let's consider the conditional we might write for the procedure to make a component more extreme.
(if (> component 127) ; If the component is greater than 127
(* component 1.2) ; Increment it by 20%
(* component 0.8)) ; Otherwise, decrement it by 20%
To turn this expression into a procedure, we need to add the define
keyword, a name (enhance-component), a lambda expression, and such.
We also want to give appropriate documentation and a bit of cleanup to the
results.
Here is the complete definition of the enhance-component procedure:
;;; Procedure:
;;; enhance-component
;;; Parameters:
;;; component, an integer
;;; Purpose:
;;; Compute an enhanced version of component. A large component gets
;;; larger. A small component gets smaller.
;;; Produces:
;;; new-component, an integer
;;; Preconditions:
;;; component is an integer between 0 and 255, inclusive
;;; Postconditions:
;;; If component > 127, then new-component is approximately 20% larger
;;; than component.
;;; If component < 128, then new-component is approximately 20% smaller
;;; than component.
(define enhance-component
(lambda (component)
(if (> component 127)
(min 255 (round (* 1.2 component)))
(round (* 0.8 component)))))
In an if expression of the form
(if test consequent alternative),
the test is always evaluated first. If its value is
#t (which means yes
or true
), then the
consequent s evaluated, and the alternative (the
expression following the consequent) is ignored. On the other hand,
if the value of the test is #f, then the consequent is
ignored and the alternative is evaluated.
Scheme accepts if expressions in which the value of the test
is non-Boolean. However, all non-Boolean values are classified as
truish
and cause the evaluation of the consequent.
It is also possible to write an if expression without the
alternative. Such an expression has the form (if test
consequent). In this case, the test is still evaluated
first. If the test holds (that is, has a value of #t or
anything other than #f), the consequent is evaluated and
its value is returned. If the test fails to hold (that is, has value
#f), the if expression has no value.
Scheme programmers tend to use the alternative-free if expression much less frequently than they use the traditional form. In general, your first inclination should be to provide both a consequent and an alternative when you write a conditional.
cond
When there are more than two alternatives, it is often more convenient
to set them out using a cond expression. Like
if, cond is a keyword. (Recall that keywords
differ from procedures in that the order of evaluation of the parameters
may differ.)
The cond keyword is followed by zero
or more lists-like expressions called cond clauses.
(cond (test-0 consequent-0) ... (test-n consequent-1) (else alternate))
The first expression within a cond clause is a test,
similar to the test in an if expression. When the value of
such a test is found to be #f, the subexpression that
follows the test is ignored and Scheme proceeds to the test at the
beginning of the next cond clause. But when a test is
evaluated and the value turns out to be true, or even truish
(that
is, anything other than #f), the consequent for that test
is evaluated and its value is the value of the whole cond expression..
Subsequent cond clauses are completely ignored.
In other words, when Scheme encounters a cond expression, it
works its way through the cond clauses, evaluating the test at
the beginning of each one, until it reaches a test that succeeds
(one that does not have #f as its value). It then makes a
ninety-degree turn and evaluates the consequent in the selected
cond clause, retaining the value of the consequent.
If all of the tests in a cond expression are found to be
false, the value of the cond expression is unspecified (that
is, it might be anything!). To prevent the surprising results that can
ensue when one computes with unspecified values, good programmers
customarily end every cond expression with a
cond clause in which the keyword else appears in
place of a test. Scheme treats such a cond clause as if it
had a test that always succeeded. If it is reached, the subexpressions
following else are evaluated, and the value of the last one
is the value of the whole cond expression.
For example, here is a cond expression that produces black,
white, or grey based only on the red component of a color, c.
(cond ((< (rgb.red c) 96) (rgb.new 0 0 0)) ((> (rgb.red c) 160) (rgb.new 255 255 255)) (else (rgb.new 127 127 127)))
The expression has three cond clauses. In the first, the test
is (< (rgb.red c) 96). If the red component of c
happens to be the small,
the value of this first test is #t, so we evaluate
whatever comes after the test to find the value of the entire expression,
in this case, the color black.
If the red component of c is not small, then we proceed instead
to the
second cond clause. Its test is (> (rgb.red c) 160), which determines if the red component is large. If it, then we return
the color white.
However, if c has a red component that is neither small nor
large,
then we proceed instead to the third cond clause.
Since the keyword else appears in this cond
clause in place of a test, we take that as an automatic success
and evaluate (rgb.new 128 128 128), so that that value of
the whole cond expression in this case is the color grey.
Although we have written our conditionals with one consequent per test (and we encourage you to do the same), it is, in fact, possible to have multiple consequents per test.
(cond (test-0 consequent-0-0 consequent-0-1 ... consequent-0-m) ... (else alternate-0 alternate-1 ... alternate-a))
In this case, when a test succeeds, each of the remaining subexpressions
(that is, consequents) in the same cond clause is evaluated
in turn, and the value of the last one becomes the value of the entire
cond expression.
and and or
As we saw in the reading on
Boolean values, both and and or provide a
type of conditional behavior. In particular, and evaluates
each argument in turn until it hits a value that is #f and
then returns #f (or returns the last value if none return
#f). Similarly, or evaluates each argument
in turn until it finds one that is not #f, in which case
it returns that value, or until it runs out of values, in which case it
returns #f.
That is, (or exp0 exp1 ... expn)
behaves much like the following cond expression,
except that the or version evaluates each expression once,
rather than twice.
(cond (exp0 exp0) (exp1 exp1) ... (expn expn) (else #f))
Similarly, (and exp0 exp1 ... expn)
behaves much like the following cond expression.
(cond ((not exp0) #f) ((not exp1) #f) ... ((not expn) #f) (else expn))
Most beginning programmers find the cond versions much more
understandable, but some advanced Scheme programmers use the
and and or forms because they find them clearer.
Certainly, the cond for and is quite repetitious.
So, what does any of this have to do with drawing? Well, we've already
seen one thing: We can use conditionals in writing color transformations.
However, there's something much more fun that we can do with conditionals:
We can use conditionals to draw simple shapes. How? We can use
region.compute-pixels! and draw one color if the pixel falls
within the shape and another if the pixel falls outside of the shape.
Here are a few conditionals that draw particular images. See if you can figure out what each one draws.
(define c0 (rgb.new 255 255 255))
(define c1 (rgb.new 255 0 255))
(define c2 (rgb.new 255 0 0))
(define c3 (rgb.new 0 255 255))
(define square (lambda (x) (* x x)))
(region.compute-pixels! canvas 0 0 199 199
(lambda (pos)
(if (< (position.row pos) (- 100 (position.col pos))) c1 c0)))
(region.compute-pixels! canvas 0 0 199 199
(lambda (pos)
(if (< (position.row pos) (+ -75 (position.col pos))) c2 c0)))
(region.compute-pixels! canvas 0 0 199 199
(lambda (pos)
(if (>= (square 50)
(+ (square (- (position.col pos) 20))
(square (- (position.row pos) 60))))
c3 c0)))
We will return to these drawings, and others, in the corresponding laboratory.
Of course, each call to region.compute-pixels! will completely
overwrite the previous drawing. What if we want to draw multiple shapes?
There are a few possibilities: The most obvious is that we can put the
different commands into a cond.
(region.compute-pixels! canvas 0 0 199 199
(lambda (pos)
(cond
((>= (square 50)
(+ (square (- (position.col pos) 20))
(square (- (position.row pos) 60))))
c3)
((< (position.row pos) (+ -75 (position.col pos))) c2)
((< (position.row pos) (- 100 (position.col pos))) c1)
(else c0))))
That works fairly well, even though it's a bit inconvenient to write. But
what if we want to draw something on top of an existing image.
In that case, we can take advantage of a special feature of
region.compute-pixels! - if the color is the special value
transparent, then
region.compute-pixels! does not modify the underlying pixel.
(We simply mention this capability here; you will take advantage of it in the
lab.)
http://www.cs.grinnell.edu/~rebelsky/Courses/CS151/History/Readings/conditionals.html.
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Reference:
[Scheme Report (R5RS)]
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Related Courses:
[CSC151.01 2007F (Davis)]
[CSC151 2007S (Rebelsky)]
[CSCS151 2005S (Stone)]
Disclaimer:
I usually create these pages on the fly
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