question on max | min op

I've read Synopsis and I wondered why to treat max and min as
operator. IMHO, view them as list functions is more reasonable. Like
below:

@test.max

is clearer than

@test[0] max @test[1]  or [max] @test.


Any reply would be really appreciated and will much help me learn
perl6. Thanks in advance!

On Tue, Apr 1, 2008 at 1:44 AM, Xiao Yafeng <xyf.xiao@gmail.com> wrote:
> I've read Synopsis and I wondered why to treat max and min as
>  operator. IMHO, view them as list functions is more reasonable. Like
>  below:
>
>  @test.max

Which is how you would probably call it in Perl6.  Or else

max(@test)


>
>  is clearer than
>
>  @test[0] max @test[1]  or [max] @test.

Which is not legal Perl6. "max" and "min" may be called "operators",
but that doesn't mean they're INFIX operator.  In Perl6, just like in
Perl5, all the builtin "functions" are really defiend as operators,
including "print" etc.  But you can always call an operator as if it
were a function/method, and in most cases you will.

pugs> [1,2,3].max
3
pugs> min(1,2,3)
1







--
Mark J. Reed <markjreed@mail.com>

On Tue, Apr 1, 2008 at 5:39 AM, Mark J. Reed <markjreed@mail.com> wrote:
>  Perl5, all the builtin "functions" are really defiend as operators,

"defined", even. (However fiendishly.)

Anyway, "function" vs "operator" is mostly a difference in terminology
that makes no difference in practice, but I believe it is technically
the "operator"-ness of these apparent functions that allows you to
call them without parentheses:

pugs> max 1, 2, 3
3




--
Mark J. Reed <markjreed@mail.com>

HaloO,

Mark J. Reed wrote:
> Anyway, "function" vs "operator" is mostly a difference in terminology

I'm not sure what the defined difference between function and operator
is in Perl 6 but I make a very clear distinction. An operator is acting
an *one* type, that is &op:(::T,T-->T) while a function is a mapping of
different types &f:(::S,::T-->::R). IOW, Operator does Function but not
the converse. So overloading an operator with inhomogenous types should
not change the abstract properties of the operator but be a convenient
form to upgrade the operants.


> that makes no difference in practice, but I believe it is technically
> the "operator"-ness of these apparent functions that allows you to
> call them without parentheses:

This is more about syntax than the distinction between operator and
function. To me max is a commutative and associative operator. That
is it could be defined in the pre-, post and infix slots.

(max 1,2,3) == (1 max 2 max 3) == (1,2,3max)
== (1,2,3\ max) == ([max] 1,2,3) == ((1,2,3)max)

Note that

(1,2,3.max) === (1,2,3 .max) === ((1,2),(3.max))

because .max tightens the precedence above that of comma and the
return type becomes a list instead of a num.


Regards, TSa.
--

The Angel of Geometry and the Devil of Algebra fight for the soul
of any mathematical being.   -- Attributed to Hermann Weyl

On Tue, Apr 01, 2008 at 05:39:36AM -0400, Mark J. Reed wrote:
> On Tue, Apr 1, 2008 at 1:44 AM, Xiao Yafeng <xyf.xiao@gmail.com> wrote: 
>
> Which is how you would probably call it in Perl6.  Or else
>
> max(@test) 
>
> Which is not legal Perl6. "max" and "min" may be called "operators",
> but that doesn't mean they're INFIX operator.

"min" and "max" are infix operators in Perl 6.  From Synopsis 3:

: * Minimum and maximum
:
:     $min0 min $min1
:     $max0 max $max1

I think they're defined as operators because of some of the
other features one can get from it, beyond just the [max] reduction:

$c = $a max $b;          # versus $c = ($a, $b).max;

$d max= $e;              # versus $d = ($d, $e).max;

@c = @a »max« @b;        # larger element of @a and @b

@e = @a »max» 100;       # each element is at least 100

Pm