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Let's consider again what happens with the triangle-recursively
function. We will find that the intermediate calculations are
deferred until all can be done.
Here is the function definition:
(defun triangle-recursively (number) "Return the sum of the numbers 1 through NUMBER inclusive. Uses recursion." (if (= number 1) ; do-again-test 1 ; then-part (+ number ; else-part (triangle-recursively ; recursive call (1- number))))) ; next-step-expression |
What happens when we call this function with a argument of 7?
The first instance of the triangle-recursively
function adds
the number 7 to the value returned by a second instance of
triangle-recursively
, an instance that has been passed an
argument of 6. That is to say, the first calculation is:
(+ 7 (triangle-recursively 6) |
The first instance of triangle-recursively
---you may want to
think of it as a little robot--cannot complete its job. It must hand
off the calculation for (triangle-recursively 6)
to a second
instance of the program, to a second robot. This second individual is
completely different from the first one; it is, in the jargon, a
`different instantiation'. Or, put another way, it is a different
robot. It is the same model as the first; it calculates triangle
numbers recursively; but it has a different serial number.
And what does (triangle-recursively 6)
return? It returns the
number 6 added to the value returned by evaluating
triangle-recursively
with an argument of 5. Using the robot
metaphor, it asks yet another robot to help it.
Now the total is:
(+ 7 6 (triangle-recursively 5) |
And what happens next?
(+ 7 6 5 (triangle-recursively 4) |
Each time triangle-recursively
is called, except for the last
time, it creates another instance of the program--another robot--and
asks it to make a calculation.
Eventually, the full addition is set up and performed:
(+ 7 6 5 4 3 2 1) |
This design for the function defers the calculation of the first step until the second can be done, and defers that until the third can be done, and so on. Each deferment means the computer must remember what is being waited on. This is not a problem when there are only a few steps, as in this example. But it can be a problem when there are more steps.
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