pipe question

[wolf volpi]

What does the pipe in this example do?

@first_name = @first_name || ''

According to the text book that this example came from, the code ensures
that an instance variable is not nil. If @first_ name is nil, @first_
name is set to the empty string.

Thank you.

 

[John Feminella]

An expression like "foo || bar" works like this: First, evaluate the
boolean value of `foo`. If it is boolean-true, then the value of this
expression is `foo`. If not, then the value of this expression is
`bar`.

This expression says, assign the current value of @first_name back
into @first_name. Alternatively, if this is the first time you're
trying to assign it, it'll be nil, so use an empty string instead.

It's equivalent to writing:

@first_name ||=3D ''

~ jf

 

[Josh Cheek]

[Note: parts of this message were removed to make it a legal post.]

What does the pipe in this example do?

@first_name = @first_name || ''

According to the text book that this example came from, the code ensures
that an instance variable is not nil. If @first_ name is nil, @first_
name is set to the empty string.

Thank you.

1) In Ruby, false and nil are treated as falsy. Everything else is treated
as truthy.

"truthy" if nil # => nil
"truthy" if false # => nil
"truthy" if true # => "truthy"
"truthy" if "str" # => "truthy"
"truthy" if 1 # => "truthy"
"truthy" if 1.23 # => "truthy"
"truthy" if /regex/ # => nil
"truthy" if :symbol # => "truthy"
"truthy" if 'r'..'ange' # => "truthy"


2) Boolean operators return the objects themselves. In an "or" (double
pipes), if the first value is falsy, the second value is returned. If the
first value is truthy, it is returned.

nil || "b" # => "b"
false || "b" # => "b"
"a" || nil # => "a"
"a" || false # => "a"
"a" || "b" # => "a"
nil || nil # => nil
nil || false # => false
false || nil # => nil
false || false # => false

3) For some reason that I don't know (probably interpreter magic) an
uninitialized variable can be referenced in a boolean equation and it will
evaluate to nil.

first_name = first_name || "" # => ""
first_name = "josh"
first_name = first_name || "" # => "josh"


So, it will set the variable to the empty string, if the variable is
undeclared, or false , or nil.

 

[wolf volpi]

Ha Ha. I was interpeting the pipes as empty arguments of an empty
block, which made no sence to me. Logical operators didn't even occure
to me. Seems funny now. Thank you John and Josh for helping me see the
light.

The || is a Logical OR Operator that returns the first true value (in
Ruby anything besides false and nil is true). I ran this demonstration
in irb:=> false
=> false
=> nil
=> 3

 

[Robert Klemme]

An expression like "foo || bar" works like this: First, evaluate the
boolean value of `foo`. If it is boolean-true, then the value of this
expression is `foo`. If not, then the value of this expression is
`bar`.

This expression says, assign the current value of @first_name back
into @first_name. Alternatively, if this is the first time you're
trying to assign it, it'll be nil, so use an empty string instead.

It's equivalent to writing:

=A0 =A0@first_name ||=3D ''

No. That is equivalent to:

@first_name || @first_name =3D ""

which is not the same as

@first_name =3D @first_name || ""

See also

irb(main):004:0> x=3DObject.new
=3D> #<Object:0x10924b1c>
irb(main):005:0> def x.[](k) puts "get";@x;end
=3D> nil
irb(main):006:0> def x.[]=3D(k,v) puts "put";@x=3Dv;end
=3D> nil
irb(main):007:0> x[1] ||=3D 2
get
put
=3D> 2
irb(main):008:0> x[1] ||=3D 2
get
=3D> 2

This has been discussed at lengths here in the past. You should be
able to find plenty of evidence.

Kind regards

robert

--=20
remember.guy do |as, often| as.you_can - without end
http://blog.rubybestpractices.com/

 

[Josh Cheek]

[Note: parts of this message were removed to make it a legal post.]

1) In Ruby, false and nil are treated as falsy. Everything else is treated
as truthy.

"truthy" if nil # => nil
"truthy" if false # => nil
"truthy" if true # => "truthy"
"truthy" if "str" # => "truthy"
"truthy" if 1 # => "truthy"
"truthy" if 1.23 # => "truthy"
"truthy" if /regex/ # => nil
"truthy" if :symbol # => "truthy"
"truthy" if 'r'..'ange' # => "truthy"

I just listed out a bunch of examples and then ran them.... But I'm looking
now, and realize that /regex/ evaluated as falsy! After some playing
around, I have decided that it is based on Perl. If you put a regex in a
variable, it treats it as a boolean "is it an object?" and behaves as I
stated previously. But if you put it in a literal, it plays Perl and checks
to see if it matches against $_

regex = /re(.)ex/
$_ # => nil

# in literal, so checks if it matches $_
# (note that it warns: regex literal in condition)
"truthy" if /re(.)ex/ # => nil

# in var, so checks if it exists
"truthy" if regex # => "truthy"

# explicitly set $_, now literal matches
$_ = "reGex"
"truthy" if /re(.)ex/ # => "truthy"

# when we look at capture group, we see the G we made
$1 # => "G"

# change the $_ to see if variable matches it
$_ = "reXex"
"truthy" if regex # => "truthy"

# nope, it should be "X" if it matched.
$1 # => "G"

# just to double check
$1 if /re(.)ex/ # => "X"


So, regex aren't nil or false, and are thus truthy. But when you throw a
regex literal in the conditional, it does some implicit operations and
doesn't pass the regex straight on through to the conditional.

 

[7stud --]

Josh Cheek wrote in post #990762:

3) For some reason that I don't know (probably interpreter magic) an
uninitialized variable can be referenced in a boolean equation and it
will
evaluate to nil.

first_name = first_name || "" # => ""
first_name = "josh"
first_name = first_name || "" # => "josh"


So, it will set the variable to the empty string, if the variable is
undeclared, or false , or nil.

I think the story goes something like this: when the parser sees any
'name =' expression, name gets entered into the symbol table.
Thereafter, you will no longer get an exception when referencing the
variable. In your code, the parser sees 'first_name =', so first_name
is entered into the symbol table, and then when ruby executes your code,
the expression on the right hand side of the equals sign is executed.

Here is an example that might prove illustrative:

if 1 > 10
x = 'hello' #never executes
end

puts x #=> nil
puts y #=> undefined local variable or method `y'
#for main:Object (NameError)

 

[Josh Cheek]

[Note: parts of this message were removed to make it a legal post.]

Josh Cheek wrote in post #990762:

I think the story goes something like this: when the parser sees any
'name =' expression, name gets entered into the symbol table.
Thereafter, you will no longer get an exception when referencing the
variable. In your code, the parser sees 'first_name =', so first_name
is entered into the symbol table, and then when ruby executes your code,
the expression on the right hand side of the equals sign is executed.

Makes sense, and passed the test I constructed to invalidate it:
lhs = rhs || "pass"
lhs # =>
# ~> -:1:in `<main>': undefined local variable or method `rhs' for
main:Object (NameError)


I probably never figured that out because it depends on the variable type,
for example:
lhs = @rhs || "pass" # => "pass"
and
lhs = $rhs || "pass" # => "pass"

Here is an example that might prove illustrative:

if 1 > 10
x = 'hello' #never executes
end

puts x #=> nil
puts y #=> undefined local variable or method `y'
#for main:Object (NameError)

Good thought. I'm honestly surprised that works! I guess it must add x to
the symbol table at parse time.

 

[Brian Candler]

Josh Cheek wrote in post #990806:

If you put a regex in a
variable, it treats it as a boolean "is it an object?" and behaves as I
stated previously. But if you put it in a literal, it plays Perl and
checks
to see if it matches against $_

Yes, this is a very ugly inheritance from Perl. Check out the flip-flop
operator too

 

[Robert Klemme]

Is that something we should be able to depend on, though -- or just an
accident of implementation?

I think it's a language feature. From the line where an assignment to
a identifier is found this identifier denotes a local variable -
whether the code is actually executed does not matter. This is only
about syntactical order. Consider

$ ruby19 -e 'puts(x) if x=3D1'
-e:1: warning: found =3D in conditional, should be =3D=3D
-e:1:in `<main>': undefined local variable or method `x' for
main:Object (NameError)
$ ruby19 -e 'x=3D1;puts(x) if x'
1
$

Kind regards

robert

--=20
remember.guy do |as, often| as.you_can - without end
http://blog.rubybestpractices.com/

 

[7stud --]

Josh Cheek wrote in post #990917:

I probably never figured that out because it depends on the variable
type,
for example:
lhs = @rhs || "pass" # => "pass"
and
lhs = $rhs || "pass" # => "pass"

Remember, by rule instance variables and global variables have a default
value of nil.

 

[Josh Cheek]

[Note: parts of this message were removed to make it a legal post.]

Josh Cheek wrote in post #990762:

I think the story goes something like this: when the parser sees any
'name =' expression, name gets entered into the symbol table.
Thereafter, you will no longer get an exception when referencing the
variable. In your code, the parser sees 'first_name =', so first_name
is entered into the symbol table, and then when ruby executes your code,
the expression on the right hand side of the equals sign is executed.

Actually, I was thinking about it, and this contradicts that a||=b expands
to a||a=b


undefined1 ||= :defined
undefined1 # => :defined

undefined2 = ( undefined2 || :defined )
undefined2 # => :defined

undefined3 || ( undefined3 = :defined )
undefined3 # =>
# ~> -:7:in `<main>': undefined local variable or method `undefined3' for
main:Object (NameError)

 

[Jesús Gabriel y Galán]

Is that something we should be able to depend on, though -- or just an
accident of implementation?

I think it's a language feature. In the Ruby Draft Specification
(http://ruby-std.netlab.jp/draft_spec/draft_ruby_spec-20091201-hl.pdf)
you can read this:

9.1.2 References to local variables
An occurrence of a local-variable-identifier can be a reference to a
local variable or a method
invocation. In order to determine whether the occurrence of a
local-variable-identifier is a
reference to a local variable or a method invocation, before the
evaluation of a local variable
scope, the scope is scanned sequentially for local-variable-identifier s.
For each occurrence of a local-variable-identifier I, take the following st=
eps:
a) If I occurs in one of the forms below, I is a reference to a local varia=
ble.
1 =88 mandatory-parameter
2 =88 optional-parameter-name
3 =88 array-parameter-name
4 =88 block-parameter-name
5 =88 variable of left-hand-side
6 =88 variable of single-variable-assignment-expression
7 =88 variable of single-variable-assignment-statement
8 =88 variable of abbreviated-variable-assignment-expression
9 =88 variable of abbreviated-variable-assignment-statement

As you can see, number 5 (variable of LHS) covers the case we are
discussing here.

Jesus.