[es] - Jako poucno i zabavno

dwarf
Beograd, Srbija

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^{01.01.2002. u 10:33 - pre 271 meseci}

Onako, da vidite kako ide perl6 i, jos bitnije, kako ide parrot. Ima par dobrih fora na koje sam se sit ismejao... :))))

http://www.panix.com/~ziggy/parrot.html#id335812

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Dragoslav Krunić

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Re: Jako poucno i zabavno

^{03.01.2002. u 22:03 - pre 271 meseci}

Sve u svemu jedan veoma interesantan dokument kao i ostali na istom sajtu koji se ticu Perla 6. Ovo mi se najvise doapda:

5. Why should I program in Parrot Assembly language?

Lots of reasons, actually. :-)

*All the cool kids are doing it.

*It's a neat hack.

*You get all of the pleasure of programming in Assembly language without any of the requisite system crashes.

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dwarf
Beograd, Srbija

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Re: Jako poucno i zabavno

^{03.01.2002. u 23:00 - pre 271 meseci}

Ne znam, meni se ona fora sa C-om definitivno najvise dojmila... :)))

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Dragoslav Krunić

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Re: Jako poucno i zabavno

^{04.01.2002. u 09:30 - pre 271 meseci}

Gledao sam nove stvari koje se uvode u Perlu 6 i primetio sam operator :=
Procitao sam opis pored njega ali mi i dalje nije jasno sta on radi. Pise da on i dalje sluzi za dodeljivanje ali ima jos jednu funkciju. Da li bi neko mogao da mi pojasni?

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dwarf
Beograd, Srbija

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Re: Jako poucno i zabavno

^{04.01.2002. u 09:52 - pre 271 meseci}

Baci gde si to video pa da pogledamo...Verovatno u RFC-ovima...Kojem tacno???

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Dragoslav Krunić

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Re: Jako poucno i zabavno

^{04.01.2002. u 11:03 - pre 271 meseci}

Razmatranje operatora nalazi se u dokumentu Apocalypse 3. Prvo sam mislio da je ovaj operator potpuno zamenio operator '=' ali prevario sam se. Operator '=' postoji i dalje. Ovaj novi isto vrsi neku dodelu vrednosti ali izgleda da nije bas tako jednostavno kao u Pascalu na primer. Evo sta tamo pise o tom operatoru:

Binary :=

We need to distinguish two different forms of assignment. The standard assignment operator, =, works just as it does Perl 5, as much as possible. That is, it tries to make it look like a value assignment. This is our cultural heritage.

But we also need an operator that works like assignment but is more definitional. If you're familiar with Prolog, you can think of it as a sort of unification operator (though without the implicit backtracking semantics). In human terms, it treats the left side as a set of formal arguments exactly as if they were in the declaration of a function, and binds a set of arguments on the right hand side as though they were being passed to a function. This is what the new := operator does. More below.

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dwarf
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Re: Jako poucno i zabavno

^{04.01.2002. u 13:48 - pre 271 meseci}

WTF??? Iskreno, ni meni nije bas najjasnije...Kao da prica o tome da to sluzi da ti prosledjujes argumente f-jama...Mozda neka nova forma closures-a???

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Dragoslav Krunić

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Re: Jako poucno i zabavno

^{04.01.2002. u 14:26 - pre 271 meseci}

I ja sam ga razumeo tako nekako i ako sam u dobro razumeo, to je onda mnogo glupa fora. Na istoj stranici, odmah ispod objasnjenja tog operatora, stoji primer u kombinaciji sa drugim operatorima ali tek mi onda nista nije jasno....

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dwarf
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Re: Jako poucno i zabavno

^{04.01.2002. u 23:08 - pre 271 meseci}

Majku mu, baci taj primer ovde pa da vidimo sta je...Posto ni meni nema nimalo smisla...

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Dragoslav Krunić

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Re: Jako poucno i zabavno

^{04.01.2002. u 23:27 - pre 271 meseci}

Primer za to je dat sa kombinacijom * operatora koji je objasnjen odmah ispod:

Unary *

Unary * is the list flattening operator. (See Ruby for prior art.) When used on an rvalue, it turns off function signature matching for the rest of the arguments, so that, for instance:

@args = (\@foo, @bar);
push *@args;

would be equivalent to:

push @foo, @bar;

In this respect, it serves as a replacement for the prototype-disabling &foo(@bar) syntax of Perl 5. That would be translated to:

foo(*@bar)

In an lvalue, the unary * indicates that subsequent array names slurp all the rest of the values. So this would swap two arrays:

(@a, @b) := (@b, @a);

whereas this would assign all the array elements of @c and @d to @a.

(*@a, @b) := (@c, @d);

An ordinary flattening list assignment:

@a = (@b, @c);

is equivalent to:

*@a := (@b, @c);

That's not the same as

@a := *(@b, @c);

which would take the first element of @b as the new definition of @a, and throw away the rest, exactly as if you passed too many arguments to a function. It could optionally be made to blow up at run time. (It can't be made to blow up at compile time, since we don't know how many elements are in @b and @c combined. There could be exactly one element, which is what the left side wants.)

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