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(emacs-lisp-intro.info)Counting function definitions


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Counting function definitions
-----------------------------

   Our immediate goal is to generate a list that tells us how many
function definitions contain fewer than 10 words and symbols, how many
contain between 10 and 19 words and symbols, how many contain between
20 and 29 words and symbols, and so on.

   With a sorted list of numbers, this is easy: count how many elements
of the list are smaller than 10, then, after moving past the numbers
just counted, count how many are smaller than 20, then, after moving
past the numbers just counted, count how many are smaller than 30, and
so on.  Each of the numbers, 10, 20, 30, 40, and the like, is one
larger than the top of that range.  We can call the list of such
numbers the `top-of-ranges' list.

   If we wished, we could generate this list automatically, but it is
simpler to write a list manually.  Here it is:

     (defvar top-of-ranges
      '(10  20  30  40  50
        60  70  80  90 100
       110 120 130 140 150
       160 170 180 190 200
       210 220 230 240 250
       260 270 280 290 300)
      "List specifying ranges for `defuns-per-range'.")

   To change the ranges, we edit this list.

   Next, we need to write the function that creates the list of the
number of definitions within each range.  Clearly, this function must
take the `sorted-lengths' and the `top-of-ranges' lists as arguments.

   The `defuns-per-range' function must do two things again and again:
it must count the number of definitions within a range specified by the
current top-of-range value; and it must shift to the next higher value
in the `top-of-ranges' list after counting the number of definitions in
the current range.  Since each of these actions is repetitive, we can
use `while' loops for the job.  One loop counts the number of
definitions in the range defined by the current top-of-range value, and
the other loop selects each of the top-of-range values in turn.

   Several entries of the `sorted-lengths' list are counted for each
range; this means that the loop for the `sorted-lengths' list will be
inside the loop for the `top-of-ranges' list, like a small gear inside
a big gear.

   The inner loop counts the number of definitions within the range.  It
is a simple counting loop of the type we have seen before.  (Note: A
loop with an incrementing counter.)  The
true-or-false test of the loop tests whether the value from the
`sorted-lengths' list is smaller than the current value of the top of
the range.  If it is, the function increments the counter and tests the
next value from the `sorted-lengths' list.

   The inner loop looks like this:

     (while LENGTH-ELEMENT-SMALLER-THAN-TOP-OF-RANGE
       (setq number-within-range (1+ number-within-range))
       (setq sorted-lengths (cdr sorted-lengths)))

   The outer loop must start with the lowest value of the
`top-of-ranges' list, and then be set to each of the succeeding higher
values in turn.  This can be done with a loop like this:

     (while top-of-ranges
       BODY-OF-LOOP...
       (setq top-of-ranges (cdr top-of-ranges)))

   Put together, the two loops look like this:

     (while top-of-ranges
     
       ;; Count the number of elements within the current range.
       (while LENGTH-ELEMENT-SMALLER-THAN-TOP-OF-RANGE
         (setq number-within-range (1+ number-within-range))
         (setq sorted-lengths (cdr sorted-lengths)))
     
       ;; Move to next range.
       (setq top-of-ranges (cdr top-of-ranges)))

   In addition, in each circuit of the outer loop, Emacs should record
the number of definitions within that range (the value of
`number-within-range') in a list.  We can use `cons' for this purpose.
(Note: `cons'.)

   The `cons' function works fine, except that the list it constructs
will contain the number of definitions for the highest range at its
beginning and the number of definitions for the lowest range at its
end.  This is because `cons' attaches new elements of the list to the
beginning of the list, and since the two loops are working their way
through the lengths' list from the lower end first, the
`defuns-per-range-list' will end up largest number first.  But we will
want to print our graph with smallest values first and the larger
later.  The solution is to reverse the order of the
`defuns-per-range-list'.  We can do this using the `nreverse' function,
which reverses the order of a list.

   For example,

     (nreverse '(1 2 3 4))

produces:

     (4 3 2 1)

   Note that the `nreverse' function is "destructive"--that is, it
changes the list to which it is applied; this contrasts with the `car'
and `cdr' functions, which are non-destructive.  In this case, we do
not want the original `defuns-per-range-list', so it does not matter
that it is destroyed.  (The `reverse' function provides a reversed copy
of a list, leaving the original list as is.)

   Put all together, the `defuns-per-range' looks like this:

     (defun defuns-per-range (sorted-lengths top-of-ranges)
       "SORTED-LENGTHS defuns in each TOP-OF-RANGES range."
       (let ((top-of-range (car top-of-ranges))
             (number-within-range 0)
             defuns-per-range-list)
     
         ;; Outer loop.
         (while top-of-ranges
     
           ;; Inner loop.
           (while (and
                   ;; Need number for numeric test.
                   (car sorted-lengths)
                   (< (car sorted-lengths) top-of-range))
     
             ;; Count number of definitions within current range.
             (setq number-within-range (1+ number-within-range))
             (setq sorted-lengths (cdr sorted-lengths)))
     
           ;; Exit inner loop but remain within outer loop.
     
           (setq defuns-per-range-list
                 (cons number-within-range defuns-per-range-list))
           (setq number-within-range 0)      ; Reset count to zero.
     
           ;; Move to next range.
           (setq top-of-ranges (cdr top-of-ranges))
           ;; Specify next top of range value.
           (setq top-of-range (car top-of-ranges)))
     
         ;; Exit outer loop and count the number of defuns larger than
         ;;   the largest top-of-range value.
         (setq defuns-per-range-list
               (cons
                (length sorted-lengths)
                defuns-per-range-list))
     
         ;; Return a list of the number of definitions within each range,
         ;;   smallest to largest.
         (nreverse defuns-per-range-list)))

The function is straightforward except for one subtle feature.  The
true-or-false test of the inner loop looks like this:

     (and (car sorted-lengths)
          (< (car sorted-lengths) top-of-range))

instead of like this:

     (< (car sorted-lengths) top-of-range)

   The purpose of the test is to determine whether the first item in the
`sorted-lengths' list is less than the value of the top of the range.

   The simple version of the test works fine unless the
`sorted-lengths' list has a `nil' value.  In that case, the `(car
sorted-lengths)' expression function returns `nil'.  The `<' function
cannot compare a number to `nil', which is an empty list, so Emacs
signals an error and stops the function from attempting to continue to
execute.

   The `sorted-lengths' list always becomes `nil' when the counter
reaches the end of the list.  This means that any attempt to use the
`defuns-per-range' function with the simple version of the test will
fail.

   We solve the problem by using the `(car sorted-lengths)' expression
in conjunction with the `and' expression.  The `(car sorted-lengths)'
expression returns a non-`nil' value so long as the list has at least
one number within it, but returns `nil' if the list is empty.  The
`and' expression first evaluates the `(car sorted-lengths)' expression,
and if it is `nil', returns false _without_ evaluating the `<'
expression.  But if the `(car sorted-lengths)' expression returns a
non-`nil' value, the `and' expression evaluates the `<' expression, and
returns that value as the value of the `and' expression.

   This way, we avoid an error.  Note: forward-paragraph, for more
information about `and'.

   Here is a short test of the `defuns-per-range' function.  First,
evaluate the expression that binds (a shortened) `top-of-ranges' list
to the list of values, then evaluate the expression for binding the
`sorted-lengths' list, and then evaluate the `defuns-per-range'
function.

     ;; (Shorter list than we will use later.)
     (setq top-of-ranges
      '(110 120 130 140 150
        160 170 180 190 200))
     
     (setq sorted-lengths
           '(85 86 110 116 122 129 154 176 179 200 265 300 300))
     
     (defuns-per-range sorted-lengths top-of-ranges)

The list returned looks like this:

     (2 2 2 0 0 1 0 2 0 0 4)

Indeed, there are two elements of the `sorted-lengths' list smaller
than 110, two elements between 110 and 119, two elements between 120
and 129, and so on.  There are four elements with a value of 200 or
larger.


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