#2 - Binary operators
Thanks to everybody helping me to discuss Pif.
We have seen some really nice polymorphic functions as EWD/...,
Pif, State Machine.
The discussion I started was meant to underline the Rebol
capability of having functions taking unlimited number of
arguments through the use of a one block argument, whose contents
are interpreted somehow.
But, I think, that the functions with a limited number of
arguments may be even more polymorphic and more interesting.
Let's try a "real world" (just kidding 8^) example:
the task is to solve the equation:
x + a = b
, where A and B are given and X is unknown:
solve: func [
{Solve the equation x + a = b}
a b
] [
b - a
]
The question is: Is this solution polymorphic?
Answer #1: Yes, you can solve the cases like:
solve 1 10
solve exp 1 pi
solve #"x" #"y"
solve 16:26 19:00
solve $1 $1'000'000
solve 25/1/2000 31/1/2000
OTOH, the answer may be:
Answer #2: No, if we consider the following case:
complex!: make object! [
re: 0
im: 0
]
solve make complex! [re: 0 im: 0] make complex! [re: 1 im: 2]
, because I didn't find other way to accomplish this than to
rewrite Solve.
If we used a slightly different version of Solve from the start
like this:
solve: func [
{Solve the equation x + a = b}
a b
] [
subtract b a
]
, then it would have been much more polymorphic.
Now the only thing that is needed is to define the following:
complex?: func [x [any-type!]] [
all [
object? x
in x 're
in x 'im
number? x/re
number? x/im
]
]
_subtract: :subtract
subtract: func [
{Returns the second value subtracted from the first.}
value1 [any-type!]
value2 [any-type!]
] [
either all [complex? value1 complex? value2] [
make complex! [
re: value1/re - value2/re
im: value1/im - value2/im
]
] [
_subtract value1 value2
]
]
I do like this kind of polymorphism. It can be observed only when
using the prefix binary operators.
OTOH, I think, that there is an even better way, how to arrange
the things.
CU later,
Ladislav
