Please explain nuances of ||=

I am reading some of the ruby files in rails and I an seeing the ||=
method used a lot.
knowing ruby the way I do, I realize that she has lots of magical
surprises and I really want to get to know this girl!

Consider the initialize method of resources.rb

51 def initialize(entities, options)
52 @plural ||= entities
53 @singular ||= options[:singular] ||
plural.to_s.singularize
54 @path_segment = options.delete(:as) || @plural
55
56 @options = options
57
58 arrange_actions
59 add_default_actions
60 set_prefixes
61 end

(OK, so one iddy bitty part of this is a rails question, and dangit,
this is a ruby forum, but you guys are smarter and have better
haircuts)

Question (1)
After several hours of searching I find virtually nothing that fully
explains the line
@singular ||= options[:singular] || plural.to_s.singularize

I think know what is does, It assigns @singular a value for a
“singular named” controller if the options hash contains the
symbol :singular. I am looking more for a full description of how ||=
and it’s friends like &&= actually work.

my current understanding shows me:
@singular = (options[:singular]) or (plural.to_s.singularize)

but the full logic seems to be more like:
if options contains a symbol named :singular then
@singular = plural.to_s.singularize
end if

But what is the value of @singular if there is no :singular symbol?
nil?

What I think is going on is a lot more like an IF statement than a OR
statement.

Question (2)
in the line: @plural ||= entities
if entities is (false?, nil? something that fails “OR”) what happens
to @plural?

Once again, the ||= seems more like an “equals If” statement rather
than an “equals Or” statement.

Question (3)
@path_segment = options.delete(:as) || @plural
So @path_segment is assigned either the return from the delete
operation (:as) ? or the plural name of the controller. Is that
correct? Where in a common resourceful routes mapping is there an :as?

Thanks in advance for any help.

You’re close on some, right on some.

@plural||=entries
simply says, if @plural is nil or false (someone correct me if I’m a
tad off), set it to entries, otherwise don’t do anything.

For @singular, close but… you need the or
@singular ||= ((options[:singular]) or (plural.to_s.singularize))

@path_segment is a little bit stranger, but for me it harkens back to c
days.
if options.delete(:as) returns any value that evalutes to true
(anything but nil or false), @path_segment is set to that value. If
options.delete(:as) fails or returns nil, @path_segment is set to
@plural.

Does that help?

On 2008-05-01, Ruby F. [email protected] wrote:

I am reading some of the ruby files in rails and I an seeing the ||=
method used a lot.

There’s a nice discussion at
URL:http://talklikeaduck.denhaven2.com/articles/2008/04/26/x-y-redux.

Regards,

Jeremy H.

Hi –

On Fri, 2 May 2008, Ruby F. wrote:

I am reading some of the ruby files in rails and I an seeing the ||=
method used a lot.
knowing ruby the way I do, I realize that she has lots of magical
surprises and I really want to get to know this girl!

x ||= y means: x || x = y

The difference is that x ||= y won’t complain if x is undefined,
whereas if you type x || x = y and there’s no x in scope, it will. The
basic idea is that if x is (a) undefined (b) nil © false, you want
to set it to y. Otherwise, you want to leave it alone. The whole
expression evaluates to x, after the assignment has happened (or not).

See http://dablog.rubypal.com/2008/3/25/a-short-circuit-edge-case for
more.

David

Hi –

On Fri, 2 May 2008, Ruby F. wrote:

I am reading some of the ruby files in rails and I an seeing the ||=
method used a lot.
knowing ruby the way I do, I realize that she has lots of magical
surprises and I really want to get to know this girl!

x ||= y means: x || x = y

No, it means: x = x || y

The difference is that x ||= y won’t complain if x is undefined,

That’s because = does the same.

mfg, simon … l

Yes, Thank you,

That is very helpful.

Google seems to puke on “||=” so I wasn’t getting any search results.

On 01.05.2008 20:54, Simon K. wrote:

x ||= y means: x || x = y

No, it means: x = x || y

I believe you are wrong.

irb(main):001:0> h={}
=> {}
irb(main):002:0> h=Hash.new true
=> {}
irb(main):003:0> h[1]
=> true
irb(main):004:0> h[1] ||= 10
=> true
irb(main):005:0> h
=> {}
irb(main):006:0>

If you were right, h would look differently:

irb(main):008:0> h=Hash.new true
=> {}
irb(main):009:0> h[1] = h[1] || 10
=> true
irb(main):010:0> h
=> {1=>true}

The same topic has been discussed exhaustively a few days ago.

Cheers

robert

Hi –

On Fri, 2 May 2008, Simon K. wrote:

x ||= y means: x || x = y

No, it means: x = x || y

This is becoming a bit of a perma-thread :slight_smile:

See Robert K.'s answer, and also have a look at the blog post I
mentioned in my response. It’s about exactly why x = x || y is not the
right expansion for x ||= y.

David

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David A. Black wrote:

|> No, it means: x = x || y
|
| This is becoming a bit of a perma-thread :slight_smile:
|
| See Robert K.'s answer, and also have a look at the blog post I
| mentioned in my response. It’s about exactly why x = x || y is not the
| right expansion for x ||= y.

x ||= y expands into x = x || y, since x == 1 expands to x = x + 1, and
we have no reason to expect ||= to behave differently (MPLOS).

Also, it works as expected if the expansion x = x || y is correct. After
all:

‘The or and || operators evaluate their first operand. If true, the
expression returns the value of their first operand; otherwise, the
expression returns the value of the second operand.’
Programming Ruby, 2nd Edition (p. 326).

And Robert’s answer works the way it works because the Hash has a
default value set, which doesn’t equate to false or nil. If you override
the default value by setting the value to false, the assignment works
again as expected.


Phillip G.
Twitter: twitter.com/cynicalryan
Blog: http://justarubyist.blogspot.com

Zero raised to the nth power remains zero.
~ – Pop Baslim
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HI –

On Fri, 2 May 2008, Phillip G. wrote:

|>>
| irb(main):001:0> h={}
|
|
Quoth Joshua B.:
=> {}
irb(main):007:0> h[3] = h[3] || ‘not false’
=> “not false”

The question, though, is what x ||= y expands to. Robert’s point is
that if you say it expands to x = x || y, that doesn’t account for
what happens with a hash that has a non-false default value.

Mind you, x || x = y doesn’t account (as I mentioned) for the fact
that if x isn’t defined, you can’t use that syntax. x ||= y is really
its own thing, and doesn’t expand 100% of the time to anything. But x
|| x = y, if you allow for the undefined x thing, describes all of the
behaviors, including the hash edge case.

David

Knowing ruby the way I do, I realize that she has lots of magical
surprises and I really want to get to know this girl!

Happens every time! Bring a girl into the bar and a fight breaks out.

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Robert K. wrote:
| On 01.05.2008 20:54, Simon K. wrote:
|> * David A. Black [email protected] (18:56) schrieb:
|>
|>> Hi –
|>>
|>> On Fri, 2 May 2008, Ruby F. wrote:
|>>
|>>> I am reading some of the ruby files in rails and I an seeing the ||=
|>>> method used a lot.
|>>> knowing ruby the way I do, I realize that she has lots of magical
|>>> surprises and I really want to get to know this girl!
|>> x ||= y means: x || x = y
|>
|> No, it means: x = x || y
|
| I believe you are wrong.
|
| irb(main):001:0> h={}
| => {}
| irb(main):002:0> h=Hash.new true
| => {}
| irb(main):003:0> h[1]
| => true
| irb(main):004:0> h[1] ||= 10
| => true
| irb(main):005:0> h
| => {}
| irb(main):006:0>
|
| If you were right, h would look differently:
|
| irb(main):008:0> h=Hash.new true
| => {}
| irb(main):009:0> h[1] = h[1] || 10
| => true
| irb(main):010:0> h
| => {1=>true}
|
| The same topic has been discussed exhaustively a few days ago.
|
| Cheers
|
| robert
|
|

You are wrong, too, though.

I refer you to Ruby-Talk 297145[0]:

Quoth Joshua B.:
‘The only reason that Chris’ example behaves like “x || x = stuff” is
because he’s defined a default value for the hash. If you set a default
value, than you’ll never have a keyed value be empty (i.e. nil). ’

x = x || ‘a value’ works when x is nil or false. Once the Hash has a
default value, it shouldn’t be false, much less nil.

See:

irb(main):001:0> h = Hash.new
=> {}
irb(main):002:0> h[1].nil?
=> true
irb(main):003:0> h[1] = h[1] || 10
=> 10
irb(main):004:0> h[2] = ‘not nil’
=> “not nil”
irb(main):005:0> h[2] = h[2] || ‘a value’
=> “not nil”
irb(main):006:0> h[3] = false
=> false
irb(main):007:0> h[3] = h[3] || ‘not false’
=> “not false”

[0] http://blade.nagaokaut.ac.jp/cgi-bin/scat.rb/ruby/ruby-talk/297145


Phillip G.
Twitter: twitter.com/cynicalryan
Blog: http://justarubyist.blogspot.com

~ Calvin: I’m a genius, but I’m a misunderstood genius. Hobbes:
What’s
misunderstood about you? Calvin: Nobody thinks I’m a genius. – Calvin
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Hi –

On Fri, 2 May 2008, Phillip G. wrote:

| mentioned in my response. It’s about exactly why x = x || y is not the
| right expansion for x ||= y.

x ||= y expands into x = x || y, since x == 1 expands to x = x + 1, and

I think you mean +=.

we have no reason to expect ||= to behave differently (MPLOS).

That’s really where this thread always starts, though: why doesn’t it
behave like x = x || y in every case? Anyway, the thing about MPOLS is
that you and I can’t invoke it :slight_smile: It was actually Matz who showed me
the x || x = y expansion (most recently at EuRuKo, where I talked to
him about it because I wanted to make sure the corrections I made to
my blog post had made it correct).

the default value by setting the value to false, the assignment works
again as expected.

Right, but hashes are allowed to have default values. So an
explanation that accounts for that case is better than one that
doesn’t. The point is that if you do a drop-in replacement x = x || y
you get a different result from ||=, and with x || x = y you don’t.
That means, definitively I think, that the latter is a more correct
expansion. (I’m not sure how it couldn’t mean that.)

Anyway, this is at least the third thread on this in recent months,
and I’ve written my blog post (and correction follow-up :slight_smile: about it.
I think I’m about ||='d out.

David

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David A. Black wrote:

|> x ||= y expands into x = x || y, since x == 1 expands to x = x + 1, and
|
| I think you mean +=.

Yes, indeed.

|> we have no reason to expect ||= to behave differently (MPLOS).
|
| That’s really where this thread always starts, though: why doesn’t it
| behave like x = x || y in every case? Anyway, the thing about MPOLS is
| that you and I can’t invoke it :slight_smile: It was actually Matz who showed me
| the x || x = y expansion (most recently at EuRuKo, where I talked to
| him about it because I wanted to make sure the corrections I made to
| my blog post had made it correct).

It doesn’t behave that way because classes can implement and override
behavior. Which can lead to surprising results, and sometimes
pathological cases of Monkeypatching. :wink:

Also, the MPLOS helps as a guideline, not as a hard and fast rule. :slight_smile:

|> And Robert’s answer works the way it works because the Hash has a
|> default value set, which doesn’t equate to false or nil. If you override
|> the default value by setting the value to false, the assignment works
|> again as expected.
|
| Right, but hashes are allowed to have default values. So an
| explanation that accounts for that case is better than one that
| doesn’t. The point is that if you do a drop-in replacement x = x || y
| you get a different result from ||=, and with x || x = y you don’t.
| That means, definitively I think, that the latter is a more correct
| expansion. (I’m not sure how it couldn’t mean that.)

However, a default value results in the Hash not evaluating to nil or
false. Which is required for || and OR to use their second operand
(since it otherwise uses the first operand), so x || x = y cannot work.
Unless we get into precedence issues here, I guess.

| Anyway, this is at least the third thread on this in recent months,
| and I’ve written my blog post (and correction follow-up :slight_smile: about it.
| I think I’m about ||='d out.

So am I.


Phillip G.
Twitter: twitter.com/cynicalryan
Blog: http://justarubyist.blogspot.com

~ Why do we drink cow’s milk? Who was the first guy who first looked
at
a cow and said “I think I’ll drink whatever comes out of these things
when I squeeze 'em!”? – Calvin
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David A. Black wrote:
| Hi –
|
| On Fri, 2 May 2008, Phillip G. wrote:
|
| Whoops – sorry :slight_smile:

As if. :stuck_out_tongue:

| I actually have a feeling we’re talking at cross-purposes, and that
| I’m somehow not getting something. All I’m saying is that h[x] ||=
| value is the same as h[x] || h[x] = value, for any hash h and any key
| x. That can be demonstrated easily (as Robert did) just by
| substituting one expression for the other. It’s an unusual case, but
| it’s legal. I’m not sure what would be needed beyond that to
| demonstrate that x = x || y is not a drop-in replacement for x ||= y.

Yeah, I’m taking the Bird’s Eye View on it all, essentially saying: 'it
works like this almost always, but Hash is a special case, since it has
default values [or whatever the implementation detail is, that makes
Hash work differently]).


Phillip G.
Twitter: twitter.com/cynicalryan
Blog: http://justarubyist.blogspot.com

~ “Endorsing products is the American way of expressing
individuality.”

  • -Calvin
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Hi –

On Fri, 2 May 2008, Phillip G. wrote:

As if. :stuck_out_tongue:
works like this almost always, but Hash is a special case, since it has
default values [or whatever the implementation detail is, that makes
Hash work differently]).

It’s possible to generalize it; someone in one of the threads wrote a
class that showed the same behavior, but I can’t remember the details.
I do wish it were otherwise; it’s definitely unexpected, I think.

David

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David A. Black wrote:
| Hi –
|
| It’s possible to generalize it; someone in one of the threads wrote a
| class that showed the same behavior, but I can’t remember the details.
| I do wish it were otherwise; it’s definitely unexpected, I think.

Well, it is unexpected if you forget about the default value for Hash,
for example.

So, I agree: On first sight it is unintuitive. Once you think about,
though…

As a tangent:
I’ve just checked my copy of The Ruby Way, and its ‘Training your
intuition’ sub-chapter. This gotcha isn’t part of it. :frowning:

Maybe it could be added for the next edition (unless Ruby 1.9/2.0 is the
reason for the new edition, and the ambiguity/idiosyncrasy is removed in
the new version of Ruby by then).


Phillip G.
Twitter: twitter.com/cynicalryan
Blog: http://justarubyist.blogspot.com

~ “Oops, I always forget the purpose of competition is to divide
people
into
winners and losers.” -Hobbes being sarcastic
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2008/5/2 Ruby F. [email protected]:

David A. Black wrote:

It’s possible to generalize it; someone in one of the threads wrote a
class that showed the same behavior, but I can’t remember the details.

http://blade.nagaokaut.ac.jp/cgi-bin/scat.rb/ruby/ruby-talk/297185

Hi –

On Fri, 2 May 2008, Phillip G. wrote:

However, a default value results in the Hash not evaluating to nil or
false. Which is required for || and OR to use their second operand
(since it otherwise uses the first operand), so x || x = y cannot work.
Unless we get into precedence issues here, I guess.

It depends what you mean by work :slight_smile: See Robert’s example.

| Anyway, this is at least the third thread on this in recent months,
| and I’ve written my blog post (and correction follow-up :slight_smile: about it.
| I think I’m about ||='d out.

So am I.

Whoops – sorry :slight_smile:

I actually have a feeling we’re talking at cross-purposes, and that
I’m somehow not getting something. All I’m saying is that h[x] ||=
value is the same as h[x] || h[x] = value, for any hash h and any key
x. That can be demonstrated easily (as Robert did) just by
substituting one expression for the other. It’s an unusual case, but
it’s legal. I’m not sure what would be needed beyond that to
demonstrate that x = x || y is not a drop-in replacement for x ||= y.

David

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