Hi,
Does anybody know where this term comes from?
"First-class object" means "something passable as an argument in a
function call", but I fail to see the connection with "object class"
or with "first-class airplane ticket". I just find the name a bit
strange. Also, if there are first-class objects, what would the
second-class objects or economy/tourist class objects be? :)
Just wondering,
Hung Jung
Jul 18 '05
24  6750 
In article <m8uub.235682$H S4.2034763@attb i_s01>,
Rainer Deyke <ra*****@eldwoo d.com> wrote: Aahz wrote:Rainer:
l[x * 2 + f(y)] = f(l[x * 2 + f(y)])
This statement contains an obvious redundancy that will make code
maintenance difficult. Python allows me to factor out some of the
redundancy:
index = x * 2 + f(y)
l[index] = f(l[index])
However, Python gives me no way to factor out the remaining
redundancy.
Sure it does: change the immutable to a mutable.
Not good enough.
Why not? Note that you're playing what is IMO an unfair game where you
keep changing the goalposts.
I'd rather write "l[x] = f(l[x])" with all of its redundancy than wrap
every conceivable immutable object in a mutable wrapper. Besides, I
don't *want* 'f' to change an object (which may also be referenced
elsewhere); I want it to change a binding.
Well, you're going to have to pay for what you want in some fashion;
Python's going to keep its default semantics, so you're going to need
*some* kind of wrapper.
And, really, "l[x] = f(l[x])" isn't that big of a deal. It's a bit of
redundancy that I'd rather not have, but it's not bad enough that I
feel the need to do anything about it.
<shrug> It's not a redundancy unless you're using a particular skewed
way of looking at things. If you're going to skew, you might as well
keep skewing until you're using a Pythonic mechanism.
--
Aahz (aa**@pythoncra ft.com) <*> http://www.pythoncraft.com/
Weinberg's Second Law: If builders built buildings the way programmers wrote
programs, then the first woodpecker that came along would destroy civilization.
On Mon, Nov 24, 2003 at 12:33:40PM -0500, Aahz wrote: In article <m8uub.235682$H S4.2034763@attb i_s01>,
Rainer Deyke <ra*****@eldwoo d.com> wrote:Aahz wrote:Rainer:
l[x * 2 + f(y)] = f(l[x * 2 + f(y)])
This statement contains an obvious redundancy that will make code
maintenance difficult. Python allows me to factor out some of the
redundancy:
index = x * 2 + f(y)
l[index] = f(l[index])
However, Python gives me no way to factor out the remaining
redundancy.
Sure it does: change the immutable to a mutable.
Not good enough.
Why not? Note that you're playing what is IMO an unfair game where you
keep changing the goalposts.
I'd rather write "l[x] = f(l[x])" with all of its redundancy than wrap
every conceivable immutable object in a mutable wrapper. Besides, I
don't *want* 'f' to change an object (which may also be referenced
elsewhere); I want it to change a binding.
Well, you're going to have to pay for what you want in some fashion;
Python's going to keep its default semantics, so you're going to need
*some* kind of wrapper.
Up for a new operator?
l[index] ()= f
<1.5-wink>, Jp
In article <ma************ *************** **********@pyth on.org>,
Jp Calderone <ex*****@intarw eb.us> wrote: On Mon, Nov 24, 2003 at 12:33:40PM -0500, Aahz wrote: In article <m8uub.235682$H S4.2034763@attb i_s01>,
Rainer Deyke <ra*****@eldwoo d.com> wrote:
I'd rather write "l[x] = f(l[x])" with all of its redundancy than wrap
every conceivable immutable object in a mutable wrapper. Besides, I
don't *want* 'f' to change an object (which may also be referenced
elsewhere) ; I want it to change a binding.
Well, you're going to have to pay for what you want in some fashion;
Python's going to keep its default semantics, so you're going to need
*some* kind of wrapper.
Up for a new operator?
l[index] ()= f
"Boot to the head. <thump>"
--
Aahz (aa**@pythoncra ft.com) <*> http://www.pythoncraft.com/
Weinberg's Second Law: If builders built buildings the way programmers wrote
programs, then the first woodpecker that came along would destroy civilization.
Jp Calderone wrote: Up for a new operator?
l[index] ()= f
I like it! :-)
I suppose you should be able to put extra args in, too, e.g.
l[index] (foo, 42)= f
--
Greg Ewing, Computer Science Dept,
University of Canterbury,
Christchurch, New Zealand http://www.cosc.canterbury.ac.nz/~greg
On Tue, 25 Nov 2003 15:07:22 +1300, "Greg Ewing (using news.cis.dfn.de )" <g2********@sne akemail.com> wrote: Jp Calderone wrote:
Up for a new operator?
l[index] ()= f
I like it! :-)
Me too ;-)
But I wonder if =()= wouldn't read more clearly, e.g.,
l[index] =()= f
and see below.
I suppose you should be able to put extra args in, too, e.g.
l[index] (foo, 42)= f
I presume that would imply that f had all optional args after the first.
What if you wanted to pass the update target in another position? E.g.,
l[index] =(foo, ??, 42)= f
where ?? is some kind of indicator for where to plug in the arg. I guess you
could use packing/unpacking if you had a tuple left side and several ??'s to match, e.g.,
a, l[index] =(foo, ??, 42, ??)= f
meaning
a, l[index] = f(foo, a, 42, l[index])
or course the targets could be simple as well
a, b =(foo, ??, ??, 42)= f
meaning
a, b = f(foo, a, b, 42)
Hm, I wonder about * for pack/unpack into arg tuple in this context
a, b =(foo, *??, 42)= f
maybe meaning
a, b = f(foo, (a,b), 42)
or did I get that backwards?
Sorry, can't help it ;-)
Regards,
Bengt Richter This thread has been closed and replies have been disabled. Please start a new discussion. Similar topics | |
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