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Re: comparison operators and *typos
======= At 2005-06-20, 10:06:21 Aubrey Jaffer wrote: =======
> | procedure: = z1 z2 z3 ...
> | procedure: < x1 x2 x3 ...
> | procedure: > x1 x2 x3 ...
> | procedure: <= x1 x2 x3 ...
> | procedure: >= x1 x2 x3 ...
> | These procedures return #t if their arguments are (respectively):
> | equal, monotonically increasing, monotonically decreasing,
> | monotonically nondecreasing, or monotonically nonincreasing.
> | ...
> | (= 0 -0) ==> #t
> | For any finite positive number x:
> | (< #e-1/0 -x -0 0 x 1/0)) ==> #t
> | These predicates are required to be transitive.
>A sequence cannot be both equal and monotonically increasing.
>(= -0 0) conflicts with (< -0 0).
My fault. (= -0 0) should be #f.
> | library procedure: infinite? z
>"Infinite" means not finite. R5RS has `ZERO?' but not `NONZERO?';
>`POSITIVE?', but not `NONPOSITIVE?'; `NEGATIVE?' but not `NONNEGATIVE?'
>`FINITE?' is more in keeping with R5RS procedure names.
The reason I define "infinite?" instead of "finite?" is that:
(cond ((infinite? x) ...)
((zero? x) ...)
As I understand the problem, we always use a predicator to select a
minority (with some single character) (instead of a big part that can
be further divided) form the whole.
> | library procedure: zero? z
> | library procedure: positive? x
> | library procedure: negative? x
> | library procedure: odd? n
> | library procedure: even? n
> | These numerical predicates test a number for a particular
> | property, returning #t or #f. See note above.
> | (positive? #e1/0) ==> #t
> | (negative? #e-1/0) ==> #t
> | (infinite? #e-1/0) ==> #t
> | (infinite? #e0/0) ==> #t
> | (positive? 0) ==> #f
> | (negative? -0) ==> #f
>What does (zero? -0) return?
>If (negative? -0) returns #f, and (= -0 0) returns #t, how does one
>test for `-0'?
(zero? -0) ==> #t
> | procedure: numerator q
> | procedure: denominator q
> | These procedures return the numerator or denominator of their
> | argument; the result is computed as if the argument was
> | represented as a fraction in lowest terms. The denominator is
> | always positive or zero. The denominator of 0 is defined to be
> | 1.
> | (numerator (/ 6 4)) ==> 3
> | (denominator (/ 6 4)) ==> 2
> | (denominator
> | (exact->inexact (/ 6 4))) ==> 2.0
>*| (denominator #e1/0) ==> 1
>*| (denominator #e-1/0) ==> -1
>*| (numerator #e1/0) ==> 0
>*| (numerator #e-1/0) ==> 0
>*Should numerator and denominator be swapped in the last 4 lines?
Yes. My fault.
>What does (exact? -0) return?
>What does (integer? -0) return?
>What does (rational? -0) return?
>What does (numerator -0) return?
>What does (denominator -0) return?
>What does (floor -0) return?
>What does (ceiling -0) return?
>What does (* -0 -0) return?
>What does (sqrt 0) return?
Please see the implementation.
> | procedure: - z1 z2
> | procedure: - z
> | optional procedure: - z1 z2 ...
> | procedure: / z1 z2
> | procedure: / z
> | optional procedure: / z1 z2 ...
> | With one argument, these procedures return the additive or
> | multiplicative inverse of their argument.
> | With two or more arguments:
> | (- z1 . z2) => (apply + z1 (map - z2))
> | (/ z1 . z2) => (apply * z1 (map / z2))
> | (- 0) ==> -0
> | (- -0) ==> 0
> | (- #e1/0) ==> #e-1/0
> | (- #-e1/0) ==> #e1/0
> | (- 3) ==> -3
> | (/ 0) ==> #e1/0
> | (/ -0) ==> #e-1/0
>*| (/ #e1/0) ==> #0
>*| (/ #e-1/0) ==> #-0
> | (/ 3) ==> 1/3
>*Should `==> #' be replaced with `==> #e'?
(/ #e1/0) ==> 0
(/ #e-1/0) ==> -0
> | Implementation
> | Here is my implementation, which is based on a Scheme implementation
> | that supports arbitrary-big integer arithmetic as well as exact
> | rational number computation. To avoid confusion with identifies in
> | base-Scheme, all procedures defined in this SRFI (except infinite?)
> | and prefixed with "my" or "my-". This reference implementation also
> | requires SRFI-9, SRFI-13, SRFI-16, and SRFI-23.
> | (separate file attached)
>There is no link to the implementation file.
It is at http://srfi.schemers.org/srfi-73/exact.scm
Sorry that I have made so many typos. A revised verions has been send to