On Wed, Jan 23, 2008 at 10:16:53PM -0500, Mark Schoonover wrote:
Ex. 1.5
I would think applicative order would return zero since it's going to
evaluate x first. Normal order, I'm not sure what it'll do. Trying it
with drscheme, it just keeps running forever without any end. Consumes
about 50% CPU, and eventually kicks the fan on the laptop CPU. When I
stop the process, it always stoped on (p) in the define.
Here's a little bit of a hint. Haskell, uses Normal order evaluation,
roughly. The equivalent program in haskell (same program, just different
notation).
p = p
test x y = if (x == 0) then 0 else y
And then I type the test expression into the interpreter
--> test 0 p
0
-->
It gives me a zero, whereas Scheme is going to hang, trying to "figure out
p".
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Think of it as substitution. Your test expression is:
(test 0 (p))
So, the expansion rule is to expand each argument first. Test expands out
to the procedure we write (using a notation that hasn't been covered, yet).
The 0 is already expanded, and what does the call to (p) expand to? The
definition of p is (p). This is still a function call, and needs to be
expanded.
In haskell, the call to 'p' evalues to p. If I try to evaluate 'p' in
haskell, the system hangs. However, since the evaluation doesn't happen
until values are needed, when x is zero, it never uses y, so it doesn't
matter that y wouldn't terminate if it tried to evaluate it.
Challenge question: there is a fairly easy way to simulate normal-order
evaluation in scheme, any idea what it might be?
David
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