Are arguments that are objects passed by reference?
in [JavaScript]
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From: Timo Reitz on 26 Oct 2009 17:57 Jorge wrote: >> <?php $bar = array('b'); >> >> /* Array ( [0] => b ) */ print_r($bar); >> >> /* &$a means pass-by-reference in PHP */ function foo(&$a) { $a = >> array(); } >> >> foo($bar); >> >> /* Array ( ) */ print_r($bar); ?> > > Here, inside foo, $a is a reference to the variable $bar, not a reference > to the array object ['b'] that $bar points to, therefore - inside foo- > $a= array(); is === $bar= array(); This is not true, the PHP manual explicitely states[1]: "$a =& $b; $a and $b are completely equal here. $a is not pointing to $b or vice versa. $a and $b are pointing to the same place." If $a would only contain a reference to the variable $bar, how would you explain the following behaviour? <?php $a=50; $b=&$a; unset($a); echo $b; // Output: 50 ?> Javascript does not have call by reference (it does have what seems to be called "call by sharing", which I never heard of before). [1] http://www.php.net/manual/en/language.references.whatdo.php
From: Jorge on 26 Oct 2009 21:12 On Oct 26, 10:57 pm, Timo Reitz <godsb...(a)arcor.de> wrote: > (...) > If $a would only contain a reference to the variable $bar, how would you > explain the following behaviour? > > <?php > $a=50; > $b=&$a; > unset($a); > echo $b; // Output: 50 > ?> Ok, I see. You're right, that is the case, and here's the explanation: http://www.php.net/manual/en/language.references.arent.php Whoever believes *that* to be what pass-by-reference truly means, is mistaken: "Call by reference In call-by-reference evaluation, a function receives an implicit reference to the argument, rather than a copy of its value. This typically means that the function can modify the argument- something that will be seen by its caller. Call-by-reference therefore has the advantage of greater time- and space-efficiency (since arguments do not need to be copied), as well as the potential for greater communication between a function and its caller (since the function can return information using its reference arguments), but the disadvantage that a function must often take special steps to "protect" values it wishes to pass to other functions. (...) Even among languages that don't exactly support call-by-reference, many, including C and ML, support explicit references (objects that refer to other objects), such as pointers (objects representing the memory addresses of other objects), and these can be used to effect or simulate call-by-reference (but with the complication that a function's caller must explicitly generate the reference to supply as an argument)." http://en.wikipedia.org/wiki/Pass_by_reference#Call_by_reference Here's what would have happened in C, had $b been a true reference: int a= 50; int *b= &a; printf("%d", *b); //prints 50 a= 0; printf("%d", *b); //prints 0 > Javascript does not have call by reference What I have said, and I maintain, is that JS's behaviour when passing an *object* is identical to that of C when passing an object by reference, and 100% compatible with what "pass-by-reference" has always meant. (which is *not* -seemingly- what PHP does) > (it does have what seems to be > called "call by sharing", which I never heard of before). LOL, what a mess: http://en.wikipedia.org/wiki/Pass_by_reference#Call_by_sharing But here's the first hint: "Although this term (call-by-sharing) has widespread usage in the Python community, identical semantics in other languages (...) are often described as call-by-value, where the value is implied to be a *reference* to the object." And the second hint: In call-by-reference evaluation, a function receives an implicit *reference* to the argument, rather than a copy of its value. -- Jorge.
From: Jorge on 26 Oct 2009 21:18 On Oct 27, 2:12 am, Jorge <jo...(a)jorgechamorro.com> wrote: > (...) > here's the first hint: > "Although this term (call-by-sharing) has widespread usage in the > Python community, identical semantics in other languages (...) are > often described as call-by-value, where the value is implied to be a > *reference* to the object." > > And the second: > In call-by-reference evaluation, a function receives an implicit > *reference* to the argument, rather than a copy of its value. And the third: call-by-reference === call-by-value when the value is a reference. -- Jorge.
From: GodsBoss on 27 Oct 2009 05:04 On 27 Okt., 02:12, Jorge <jorge(a)jorgechamorro.com> wrote: > On Oct 26, 10:57 pm, Timo Reitz <godsboss(a)arcor.de> wrote: > > <?php > > $a=50; > > $b=&$a; > > unset($a); > > echo $b; // Output: 50 > > ?> > > Ok, I see. You're right, that is the case, and here's the explanation: > http://www.php.net/manual/en/language.references.arent.php > > Whoever believes *that* to be what pass-by-referencetruly means, is > mistaken: I don't think so, call-by-reference is exactly what PHP does and it is not what C with its pointers does. > "Call byreference > In call-by-referenceevaluation, a function receives an implicit > reference to the argument, rather than a copy of its value. This > typically means that the function can modify the argument- something > that will be seen by its caller. > (...) > http://en.wikipedia.org/wiki/Pass_by_reference#Call_by_reference In PHP, you can do the following: // Let $a be a variable with any value. // Let doSomething be a function with call-by-reference. $b=$a; doSomething($b); // Now $a != $b ! This is exactly, what call-by-reference means! In javascript, you can't do this: // Let a be a variable with any value. // Let doSomething be a function. var b=a; doSomething(a); // a===b, in every case. If javascript supported call-by-reference, you could have changed a in a manner that a!==b (maybe you could cheat by using something like Mozilla's access to variables within closures, but it would not work in vanilla ECMAScript). > Here's what would have happened in C, had $b been a truereference: > > int a= 50; > int *b= &a; > > printf("%d", *b); //prints 50 > a= 0; > printf("%d", *b); //prints 0 In PHP, this works exactly the same: $a=50; $b=&$a; $a=0; // Now $b is 0, too! But please notice how you use variables of different types in C (an integer and a pointer to an integer), while in PHP both variables are of the same type. And, again, javascript - how would I achieve the same effect here? > > Javascript does not have call byreference > > What I have said, and I maintain, is that JS's behaviour when passing > an *object* is identical to that of C when passing an object byreference, and 100% compatible with what "pass-by-reference" has > always meant. (which is *not* -seemingly- what PHP does) I agree with the part that javascript does what C does, but C does not have real call-by-reference either, as is stated in the Wikipedia article you cited: "Even among languages that _don't exactly support call-by-reference_, many, _including C_ and ML, support explicit references (objects that refer to other objects), such as pointers (objects representing the memory addresses of other objects), and these can be used to effect or simulate call-by-reference (but with the complication that a function's caller must explicitly generate the reference to supply as an argument)." > > (it does have what seems to be > > called "call by sharing", which I never heard of before). > > LOL, what a mess:http://en.wikipedia.org/wiki/Pass_by_reference#Call_by_sharing > > But here's the first hint: > "Although this term (call-by-sharing) has widespread usage in the > Python community, identical semantics in other languages (...) are > often described as call-by-value, where the value is implied to be a > *reference* to the object." > > And the second hint: > In call-by-referenceevaluation, a function receives an implicit > *reference* to the argument, rather than a copy of its value. You have to distinguish between call-by-reference and call-by-value where the value is a reference. They are not the same.
From: Jorge on 27 Oct 2009 11:43
On Oct 27, 10:04 am, GodsBoss <godsb...(a)gmail.com> wrote: > (...) > I don't think so, call-by-reference is exactly what PHP does and it is > not what C with its pointers does. > > > "Call by reference > > In call-by-reference evaluation, a function receives an implicit > > reference to the argument, rather than a copy of its value. This > > typically means that the function can modify the argument- something > > that will be seen by its caller. > > (...) > >http://en.wikipedia.org/wiki/Pass_by_reference#Call_by_reference > > In PHP, you can do the following: > > // Let $a be a variable with any value. > // Let doSomething be a function with call-by-reference. > $b=$a; > doSomething($b); > // Now $a != $b ! > > This is exactly, what call-by-reference means! What I would have said as a C programmer is that given the outcome: // Now $a != $b, $b must have been -necessarily- passed by reference. (Assuming that neither $a nor $b were already in scope inside doSomething() in which case they could have been touched directly, or, assuming that only $b has been modified and it's been touched using no other means than doSomething()'s first argument.) > In javascript, you can't do this: > > // Let a be a variable with any value. > // Let doSomething be a function. > var b=a; > doSomething(a); > // a===b, in every case. > > If javascript supported call-by-reference, you could have changed a in > a manner that a!==b (...) Agreed. I've never said that JS supported call-by-reference, I said that saying (as in Mozilla's MDC) that "objects are passed by reference" is correct. > > > Here's what would have happened in C, had $b been a true reference: > > > int a= 50; > > int *b= &a; > > > printf("%d", *b); //prints 50 > > a= 0; > > printf("%d", *b); //prints 0 > > In PHP, this works exactly the same: > > $a=50; > $b=&$a; > $a=0; > // Now $b is 0, too! Ok. Perfect. That's what I would have expected. How (or why) is that compatible with your previous PHP example in which the behaviour was (apparently) the opposite ? > But please notice how you use variables of different types in C > (an integer and a pointer to an integer), while in PHP both variables > are of the same type. That. Yes. And everything is pretty clear and unambiguous when the referencing and de-referencing operators ["*","->","&"] are used explicitly :-) > And, again, javascript - how would I achieve the same effect here? It's not possible. > > > Javascript does not have call byreference > (...) > You have to distinguish between call-by-reference and call-by-value > where the value is a reference. They are not the same. Thanks Timo, I see your point. You've put it very well, really. Interesting and constructive. But let me ask you this: Consider: /* ****************** JavaScript: */ x= {}; //a reference to {} o= {}; //a different reference to a different {} function f (a) { a.p= 27; a= x; a.p= 33; } f(o); console.log([o.p, x.p]); // 27,33 /* ****************** C: */ typedef struct { int p; } object; object xx; //creates 1st {} object oo; //creates 2nd {} object *o= &oo; //a reference to 1st {} object *x= &xx; //a different reference to a different {} function f (object *a) { a->p= 27; //or (*a).p= 27; a= x; //or a= &xx; a->p= 33; //or (*a).p= 33; } f(o); //or f(&oo); printf("%d,%d",oo.p, xx.p); // 27,33 printf("%d,%d",o->p, x->p); // 27,33 (try to ignore the unavoidable differences in syntax, due to var typing and explicit de/referencing) Question # 1: Do you agree that o are references in both programs ? Question # 2: Do you agree that it's o's value what gets passed (by copy) to f() ? Question # 3: Do you agree that both programs produce exactly the same results ? Question # 4: What if not a reference could be 'o' in order to produce these results/behaviour ? Well. Now ask whoever C programmer you like, how is it called when 'oo' is passed like that. His answer is going to be "it's being passed by reference". Simply because it's a reference to 'oo' what's being passed. In JS just like in C, only that in C it's more evident because of the explicit use of &oo. It has always been called so. What that article in MDC does is to call it as it's been always called, when they say that "objects are passed by reference". Now, somebody may want to lucubrate about other aspects or about the entire mechanism of parameter passing in these or that dynamic languages, but not me. That's beyond this simple fact, ISTM. Cheers, -- Jorge. |