How are arguments a legit argument to Array.slice?

[lorlarz]

In the code sample below, how are arguments a legitimate
argument to Array.slice?


Function.prototype.bind = function(){
var fn = this, args = Array.prototype.slice.call(arguments), object =
args.shift();
return function(){
return fn.apply(object,
args.concat(Array.prototype.slice.call(arguments)));
};
};
var myObject = {};
function myFunction(){
return this == myObject;
}

myFunction.bind(myObject)();

 

[Richard Cornford]

lorlarz wrote

In the code sample below, how are arguments a legitimate
argument to Array.slice?

Function.prototype.bind = function(){
var fn = this, args = Array.prototype.slice.call(arguments),
object = args.shift();
return function(){
return fn.apply(object,
args.concat(Array.prototype.slice.call(arguments)));
};
};
var myObject = {};
function myFunction(){
return this == myObject;
}

myFunction.bind(myObject)();

Code that is intended to be read by humans (so anything posted to Usenet
with the intention of its being examined by other participants in a
group) should be indented (using spaces in posts not tabs as tabs don't
receive uniform (or necessarily useful) handling in newsreader
software).

In the code above - arguments - is never an argument to Array.slice. All
occurrences of - arguments - as an argument are as an argument to -
Function.prototype.call -.

Richard.

 

[RobG]

In the code sample below, how are arguments a legitimate
 argument to Array.slice?

Function.prototype.bind = function(){
var fn = this, args = Array.prototype.slice.call(arguments), object =
args.shift();
return function(){
return fn.apply(object,
args.concat(Array.prototype.slice.call(arguments)));};

Seems to me that:

return fn.apply(object, args);

is sufficient.

 

[lorlarz]

lorlarz wrote








Code that is intended to be read by humans (so anything posted to Usenet
with the intention of its being examined by other participants in a
group) should be indented (using spaces in posts not tabs as tabs don't
receive uniform (or necessarily useful) handling in newsreader
software).

In the code above - arguments - is never an argument to Array.slice. All
occurrences of - arguments - as an argument are as an argument to -
Function.prototype.call -.

Richard.- Hide quoted text -

- Show quoted text -

This code is from a draft of a new JavaScript book
by a major expert. He is grabbing
the arguments property of the function and passing
it as an argument to Array.prototype.slice

The contents of the arguments property that the function
receives is just the object myObject, so there seems to be
nothing to slice.

I may have to ask the author himself; perhaps this
is a code error in the draft.

 

[lorlarz]

Seems to me that:

  return fn.apply(object, args);

is sufficient.

Yes, what you say at least makes sense to me.

 

[Richard Cornford]

^^^^^^^^^^^^^^^^^^^^^

I did _not_ write that final text. Why are you attributing words to me
that I did not write? That represents a serious breach of netiquette,
and is disingenuous at the very least.

This code is from a draft of a new JavaScript book

And the code is indented in that presentation (that is not a question,
and I don't need to see it in order to *know* that it *is* indented).

by a major expert.
YMMV

He is grabbing

What it the technical meaning of "grabbing" in javascript?

the arguments property of the function and passing
it as an argument to Array.prototype.slice

No he is not. That is not what the code above does, as I have already
told you once; at no point in the code above does an - arguments -
object get used as an argument to the - slice - method.

The contents of the arguments property that the function
receives is just the object myObject, so there seems to be
nothing to slice.

The - slice - method is used in order to transfer the values of the
'array index' properties of an - arguments - object to corresponding
properties of a newly created Array object. This allows - shift - to be
called on that Array in order to extract its first element (the
reference to the - myObject - object) and leave any other values for
later use in any call to the - fn - function.

I may have to ask the author himself;

You still may have to work out what the correct question is.

perhaps this is a code error in the draft.

No, the code is fine; a straight reproduction of a run-of-the-mill -
bind - implementation. Obviously the accompanying text is not very
effective as an explanation of what it going on here, but that is 'major
experts' for you.

Richard.

 

[lorlarz]

           ^^^^^^^^^^^^^^^^^^^^>> - Show quoted text -

   ^^^^^^^^^^^^^^^^^^^^^

I did _not_ write that final text. Why are you attributing words to me
that I did not write? That represents a serious breach of netiquette,
and is disingenuous at the very least.


And the code is indented in that presentation (that is not a question,
and I don't need to see it in order to *know* that it *is* indented).


What it the technical meaning of "grabbing" in javascript?


No he is not. That is not what the code above does, as I have already
told you once; at no point in the code above does an - arguments -
object get used as an argument to the - slice - method.


The - slice - method is used in order to transfer the values of the
'array index' properties of an - arguments - object to corresponding
properties of a newly created Array object. This allows - shift - to be
called on that Array in order to extract its first element (the
reference to the - myObject - object) and leave any other values for
later use in any call to the - fn - function.


You still may have to work out what the correct question is.


No, the code is fine; a straight reproduction of a run-of-the-mill -
bind - implementation. Obviously the accompanying text is not very
effective as an explanation of what it going on here, but that is 'major
experts' for you.

Richard.- Hide quoted text -

- Show quoted text -

You were automatically quoted by google and it looks correct to
me. There is just a single ">" in front of your lines and
">>" in front of where you quote me.

In any case, I simple took the google quote and did nothing
but reply below. If there is any validity to your
concern, you will have to take it up with google.

I am trying, but do not yet quite understand your
explanation. Could you spell it out bit by bit
more?

I agree with you that the expert did indeed
fail to explain the code, becuase he provides
no explanation. You may have done a bit better,
but I am still unclear. I think you may be able
to bring me "over the line" if you just detail
a bit more what happens.

(As with my last reply, I used the google
quotinog of your post and did nothing
to alter the way it was quoted.)

 

[lorlarz]

You were automatically quoted by google and it looks correct to
me.   There is just a single ">" in front of your lines and
">>" in front of where you quote me.

In any case, I simple took the google quote and did nothing
but reply below.  If there is any validity to your
concern, you will have to take it up with google.

I am trying, but do not yet quite understand your
explanation.  Could you spell it out bit by bit
more?

I agree with you that the expert did indeed
fail to explain the code, becuase he provides
no explanation.  You may have done a bit better,
but I am still unclear.  I think you may be able
to bring me "over the line" if you just detail
a bit more what happens.

(As with my last reply, I used the google
 quotinog of your post and did nothing
 to alter the way it was quoted.)- Hide quoted text -

- Show quoted text -

P.S. I do see what google is doing to you, Cornford,
and it just did it to me ABOVE. I will make sure
I make up for the terrible thing google does
ALTHOUGH THAT "quoted text" IS ABSOLUTELY EMPTY
and only a weird person would worry or take offense.

No offense.

Still, again, Cornford, I would be most appreciative
if you could detail you last response (if possible).

 

[lorlarz]

           ^^^^^^^^^^^^^^^^^^^^>> - Show quoted text -

   ^^^^^^^^^^^^^^^^^^^^^

I did _not_ write that final text. Why are you attributing words to me
that I did not write? That represents a serious breach of netiquette,
and is disingenuous at the very least.


And the code is indented in that presentation (that is not a question,
and I don't need to see it in order to *know* that it *is* indented).


What it the technical meaning of "grabbing" in javascript?


No he is not. That is not what the code above does, as I have already
told you once; at no point in the code above does an - arguments -
object get used as an argument to the - slice - method.


The - slice - method is used in order to transfer the values of the
'array index' properties of an - arguments - object to corresponding
properties of a newly created Array object. This allows - shift - to be
called on that Array in order to extract its first element (the
reference to the - myObject - object) and leave any other values for
later use in any call to the - fn - function.


You still may have to work out what the correct question is.


No, the code is fine; a straight reproduction of a run-of-the-mill -
bind - implementation. Obviously the accompanying text is not very
effective as an explanation of what it going on here, but that is 'major
experts' for you.

Richard.

Let me see if I can spell the situation out for myself.
The call method looks for a real array and thus converts
the arguments object into a real array and then sends
the first argument of that resultant array
to the call method as the context
for "this" and IF there where any more arguments
(which in this example there are not), they would
be passed to the called function (Array.slice) IN
that context.

If this is correct, I guess my only remaining question
is what is the word "prototype" in this statement
var args = Array.prototype.slice.call(arguments); ??

I am doing all I can for myself. Can you
bring me "over the line" , Richard??

 

[Lasse Reichstein Nielsen]

You were automatically quoted by google and it looks correct to
me. There is just a single ">" in front of your lines and
">>" in front of where you quote me.

There is also the text "- Hide quoted text -" and "- Show quoted
text", which are attributed to Richard, and that wasn't there in his
message.

In any case, I simple took the google quote and did nothing
but reply below. If there is any validity to your
concern, you will have to take it up with google.

Uhm, no. You are responsible for the tools you chose to use.
If Google Groups doesn't work satisfactory, you should take
it up with Google.

I am trying, but do not yet quite understand your
explanation. Could you spell it out bit by bit
more?

I'm jumping into the middle here, but was the problem the
interpretation of
Array.prototype.slice.call(arguments)

This calls the "call" method on the function "Array.prototype.slice"
with the argument "arguments".

This gives approximatly the same effect as placing the function
as a method of the arguments object and calling it.

arguments.somename = Array.prototype.slice;
... arguments.somename();

just without actualy creating a property on the arguments object.

(As with my last reply, I used the google
quotinog of your post and did nothing
to alter the way it was quoted.)

And it again introduced spurious text. Maybe you will have
to do something to avoid this, since nothing apparently
doesn't do the job

 

[Lasse Reichstein Nielsen]

(you can trim your quotings a lot more without losing context)

Let me see if I can spell the situation out for myself.
The call method looks for a real array

No. The call method looks for an object and, optionally, some more
arguments.

and thus converts
the arguments object into a real array and then sends
the first argument of that resultant array
to the call method as the context
for "this" and IF there where any more arguments
(which in this example there are not), they would
be passed to the called function (Array.slice) IN
that context.

The call method the calls the "Array.prototype.slice" function with
"this" set to that object. I.e., when calling "slice", it is as if it
was a method on the arguments object.

The slice function is generic, and doesn't need an array as its "this"
value, just something with a "length" property.
The slice function then creates a new array that contains the same
elements as its "this" value. This converts the arguments object
to an array (or rather, creates a new array).

If this is correct, I guess my only remaining question
is what is the word "prototype" in this statement
var args = Array.prototype.slice.call(arguments); ??

Array.prototype.slice references a function. This is the same function
that Array objects inherit through their prototype chain. I.e., if
var a = [];
is an array, then Array.prototype.slice and a.slice is the same function.

/L

 

[Lasse Reichstein Nielsen]

Seems to me that:

return fn.apply(object, args);

is sufficient.

The above code allows partial application. I.e.,

function foo(a,b,c) { /*...*/ }
var foob = foo.bind(object, 42);
foob(37,"doh"); // equvialent to object.call(foo,42,37,"doh")

The conversion from arguments-object to array is probably used because
the "concat" function requires an array as argument, and preferably an
array to work on (although it might work on an arguments object).

/L

 

[lorlarz]

I'm jumping into the middle here, but was the problem the
interpretation of
 Array.prototype.slice.call(arguments)

This calls the "call" method on the function "Array.prototype.slice"
with the argument "arguments".

This gives approximatly the same effect as placing the function
as a method of the arguments object and calling it.

 arguments.somename = Array.prototype.slice;
 ... arguments.somename();

just without actualy creating a property on the arguments object.
[snip]

Believe it or not, I think I understand. The function is
being called with only one argument (the arguments object)
and thus that is the context of the function call on
the slice method. Did I say that correctly?

 

[lorlarz]

On Aug 29, 2:48 pm, Lasse Reichstein Nielsen <[email protected]>
wrote:
[snip]

Array.prototype.slice references a function. This is the same function
that Array objects inherit through their prototype chain. I.e., if
 var a = [];
is an array, then Array.prototype.slice and a.slice is the same function.

/L

Again, I believe I understand. If you are not
dealing with a specific array, you need to
call Array.prototype to use the slice method.

I am still not clear on why Array.slice would not
work though. Have an old Java background
that might be causing confusion.

 

[lorlarz]

The above code allows partial application. I.e.,

 function foo(a,b,c) { /*...*/ }
 var foob = foo.bind(object, 42);
 foob(37,"doh"); // equvialent to object.call(foo,42,37,"doh")

The conversion from arguments-object to array is probably used because
the "concat" function requires an array as argument, and preferably an
array to work on (although it might work on an arguments object).

/L

Indeed partial application is exactly the context the
code came up in. I bow to your expertise. You certainly
know what you are talking about. You surely
have my respect.

I will strive to make up for google's short-comings
in responding to other. Best wishes.

 

[dhtml]

lorlarz wrote

[snip]


This code is from a draft of a new JavaScript book
by a major expert.

LOL. I think I know whose book this is.

He is grabbing

the arguments property of the function and passing
it as an argument to Array.prototype.slice

Is he using his sword?

The contents of the arguments property that the function
receives is just the object myObject, so there seems to be
nothing to slice.

I think I see Lasse is helping you just a little too much. I've
mentioned the EcmaScript specification to you before. Do a little
research there first. You can search that specification for "call(" .
Then search for "slice(" and see what that does.

You can find the "unofficial edition" spec in HTML on bclary.com and the
official version in pdf here:
http://www.ecma-international.org/publications/standards/Ecma-262.htm

Garrett

 

[lorlarz]

lorlarz wrote
In the code sample below, how are arguments a legitimate
argument to Array.slice?

[snip]



This code is from a draft of a new JavaScript book
by a major expert.  

LOL. I think I know whose book this is.

He is grabbing

the arguments property of the function and passing
it as an argument to Array.prototype.slice

Is he using his sword?

The contents of the arguments property that the function
receives is just the object myObject, so there seems to be
nothing to slice.

I think I see Lasse is helping you just a little too much.  I've
mentioned the EcmaScript specification to you before. Do a little
research there first. You can search that specification for "call(" .
Then search for "slice(" and see what that does.

You can find the "unofficial edition" spec in HTML on bclary.com and the
official version in pdf here:http://www.ecma-international.org/publications/standards/Ecma-262.htm

Garrett

I really do not consider you to have a legitimate criticism
of me asking the question. IF THIS rather hard-to-understand
routine is not something one can ask about here, WHAT IS??
One can _always_ find _all_ the answers elsewhere (FACT), but
can sometimes learn more more clearly from some of the better
people HERE. And, that is what the group is about. (Otherwise,
if someone can always find the answer elsewhere & in _your view_
should not ask the question here, NO QUESTIONS would be
"legitimate" here!!!! Isn't this silly???)

Now, how is this particular thing is harder than just call and
slice? Well, first you must know that arguments is an object
and that an object is seen as a single entity. Then you must
realize that the first argument to call simply sets the
context of "this" plus that slice is just being using to make
this object a one element array. This is not something all
readily understand.

Truth is I have a remaining question . It is that in
the code,
return fn.apply(object,
args.concat(Array.prototype.slice.call(arguments)));
,
it seems apply (there) is potentially getting an extra,
array element (because the first element in arguments
is still the object (the one setting the context of "this).

Thus, I am not sure I still fully understand and
am trying not to be embarrassed asking the question
HERE, despite criticism of asking the question I
aready have.

So, Lasse, I am still listening.