what does `![var_name]` mean in Ruby? - ruby

I am dealing with someone else's code base, and I am trying to decipher the following line:
if !foo #<~ this line
...
end
I noticed that foo is supposed to return false, but with this bang! method it returns true. Does this mean that this ![var] business is similar to saying:
if foo != nil

Yes, the above code means to run the code if the condition (foo) is false. I won't say its similar to foo != nil, its more like !(foo) where foo can be nil or false.
You can write
if !foo #<~ this line
...
end
as
unless foo
...
end
But as per Ruby style guide, you should use unless, only if the code can be represented in a single line.
do_something unless foo

It means "if not foo", and it effectively returns the opposite of foo. So if foo = true, !foo returns false, and if foo = false, !foo returns true.

!foo
is syntactic sugar for
foo.!
i.e. sending the message ! to the object returned by evaluating the expression foo.
The method ! is defined in BasicObject, and it is documented just like any other method, even if rather tersely:
!obj → true or false
Boolean negate.

In this case the ! negates the value that follows it. In Ruby, only false and nil are considered falsy values, or in other words, values that will be considered false in a conditional statement like if. All other values (true, 1, "some-string", etc...) are considered truthy (values that will be considered true in a conditional).
So !false, is true. !nil is true. !true is false.

Related

Why does an empty String returns false when comparing to boolean in ruby?

It might be a dumb question, but can someone please explain this:
if ""
true
else
false
end
=> true (OK)
!!("")
=> true (OK)
"" === true
=> false (OK)
"" == true
=> false (But why?)
Have a nice day!
The basic concept of this question is a deep misunderstanding of Ruby operators. Here's the short of it - there's no such thing as operators in Ruby! All this !, =, == and === that you throw around - these are not operators.
So what's going on?
Ruby is an object oriented language (a real one, not like that fake Java and JavaScript things) and all these fancy character sequences you think of as operators are actually method calls on the object in their left:
== method usually checks for equality - is the object on the right equal in content.
=== method is not used for "strict equality" or "identity" like it is often in other languages - some Ruby objects that implement it use it for "membership" tests, like in ranges ((1..3) === 2 ==> true) or regular expressions (/el/ === "hello" ==> true you can think of regular expressions as the group of all strings that would match), others implement it as equals.
So - how is the if working? Well, if and other "forced boolean contexts" check for "falsehood". In Ruby we recognize two false values - false and nil, everything else will run the truth branch of an if. This is actually the method ! that you also used in your second example. This method is implemented in BasicObject to return true for all objects, except if the object's type is the type of false - FalseClass or the type for nil - NilClass, in which case it returns true.
So your examples actually mean:
Check for the truthiness of "". Because that value's type is neither FalseClass or NilClass, it is always true.
For the value "", call the method ! - which will return false because the object is neither a FalseClass or a NilClass - then call ! on the result, which will return true because false is a FalseClass instance.
For the value "" call the method === with the argument true, but since it is an alias for == - call that instead (see 4).
For the value "" call the method == with the argument true. String's implementation of == will never return true for an argument that isn't a String type.
The only falsy values in Ruby are false and nil. Everything else is "truthy," but not necessarily equal to true. This includes empty objects like a String, Hash, or Array.
Pragmatically, it might help to think of == as "comparably equivalent to" rather than "equals." For example:
1 == 1.0 #=> true
This is true even though one is an Integer and one is a Float because they are comparably equivalent in value, even if they aren't the same object or of the same type.
In the same way, "" is truthy because it is not comparably equivalent to false or nil. However, it's also not the same object type as true, nor comparably equivalent to true. An empty String is simply "not falsy," which makes it truthy but not actually true.
Remember, only false and nil are falsy. Everything else, and I mean everything, is truthy even if it isn't strictly speaking true.
In Ruby, the only "falsey types" are FalseClass and NilClass, which have the instances false and nil respectively. All other values are considered "truthy". This is possibly different to what you'd expect coming from other C-like values, in which we do things like this pretty freely:
int x = get_value();
if (x) { /* implied x != 0 }
So, if you had something like this in Ruby:
puts 0 if 0 # => "0"
puts 1 if "" # => "1"
puts 2 if [] # => "2"
puts 3 if false # => nil
puts 4 if true # => "4"
puts 5 if nil # => "5"
So, if "" acts truthy, why isn't it equal to true? If that were how we defined ==, then this would also need to resolve to true then, since both values are truthy:
"1" == "2"
The difference here is that == is asking if two things are the same, which "" and true are most certainly not. Further, Ruby does not automatically convert types for you (like other languages like JavaScript do), so "" does not automatically get converted to a boolean during its comparison.

Why does Ruby expression with double ampersand using a return statement cause a syntax error

def foo
true && return false
end
def bar
true and return false
end
foo method causes a syntax error. bar doesn't. Why?
Assuming I want to evaluate one-liners with a return statement (similar to a certain commonly used shell/bash expression), would this be a recommended way to do it, or is there a more recommended approach?
By operator associativity strength,
true && return false
evaluates to
(true && return) false
of which
true && return
is valid, but the false after it is invalid. You cannot have two statements lined up without anything in between.
Side Note
It is worth noting that and and && are not equivalent.
and is a flow control operator while && is a Boolean operator. What is the difference?
One example of the differences, using and, you can assign values.
value = nil and value / 42
This would fail if you tried it as below
value = nil && value / 42
Original Question 1
Assuming I want to evaluate one-liners with a return statement (similar to a certain > commonly used shell/bash expression), would this be a recommended way to do it, or is there > a more recommended approach?
The way I have always seen this done in Ruby is this:
value if conditional
This will return the value if the conditional is met and nil otherwise. No need for the return statement if this is the last command in the function!
If you are using this for an early exit to a function, I prefer using an unless. For instance...
return unless arguments_are_valid
Original Question 2
foo method causes a syntax error. bar doesn't. Why?
It's a matter of operator precedence. See the example below showing how they are evaluated.
(true && return) false
(true) and (return false)
Because of the && operator precedence, the following line
true && return false
evaluates as
(true && return) false
that does not makes sense. In order to execute your code you need to use
def foo
true && (return false)
end
and doesn't suffer of the same issue because and has lower precedence than &&.
if there is need for shorten statements use
def foo
return false if true
end
def bar
return false if true
end
return is not a function :) therefore it doesn't make sense to say
when true and return is ok send a false

Why do method declarations evaluate to nil in ruby?

Defining a method doesn't seem to evaluate to a truthy value as can be checked by putting one inside an if condition:
if(def some_method; puts "random text"; end) then
puts "declaration evaluates to true"
else
puts "declaration evaluates to false"
end
# => declaration evaluates to false
Why/How does a method declaration evaluate to nil?
It actually evaluates to nil. This makes sense; why would a method creation return anything?
irb(main):001:0> def test; print 'test'; end
=> nil
However, it has to return something, so to return "nothing" would be to return nil.
Every statement in Ruby evaluates to something. The def statement's value is not supposed to be checked and is therefore nil.
You will find the behavior you are looking for in the reflective "meta-programming" method define_method.
class EmptyClass
m = define_method(:newmethod) {p "I am the new method"}
p m # => <Proc:0x50b3f359#E:\NetBeansProjects\RubyApplication1\lib\main.rb:6>
end
From Ruby gotchas:
Boolean evaluation of non-boolean data is strict: 0, "" and [] are all evaluated to true. In C, the expression 0 ? 1 : 0 evaluates to 0 (i.e. false). In Ruby, however, it yields 1, as all numbers evaluate to true; only nil and false evaluate to false. A corollary to this rule is that Ruby methods by convention — for example, regular-expression searches — return numbers, strings, lists, or other non-false values on success, but nil on failure. This convention is also used in Smalltalk, where only the special objects true and false can be used in a boolean expression.
Method definions such as def some_method; puts "random text"; end always return nil.
Now, that means the method is evaluated to nil. According to the Ruby Documentation:
Returns false if obj is nil or false; true otherwise.
Since your method return nil, if will evaluate it as false therefore execute the else statement.

Double ampersand in Ruby

I am using the authlogic gem with Ruby on Rails, and I have been using the following to obtain the id of the user that is currently logged in:
current_user = UserSession.find
id = current_user && current_user.record.id
I'm not understanding how current_user && current_user.record.id returns the current user id. I would think this would return a boolean. Can someone explain how this works?
There is no Boolean type in Ruby; Ruby has a rather simple view of truth (or more precisely, it has a rather simple view of falsehood).
the false object, which is the singleton instance of FalseClass is considered falsy
the nil object, which is the singleton instance of NilClass is falsy
every other object is truthy (including, obviously, the true object, which is the singleton instance of TrueClass)
[BTW: this means that a lot of objects that are considered falsy in some other languages, are actually truthy in Ruby, like the integer 0, the real value 0.0, the empty string, the empty array, the empty hash, the character 'F']
So, the Boolean operators &&, ||, and and or do not return Boolean values. Instead they return the first object that determines the outcome of the expression.
(They are also short-circuiting, which means that they only evaluate the minimum sub-expressions that are needed to determine the outcome of the expression. So, an alternate formulation would be that they return the result of the last expression that was evaluated. Which, in turn, is analogous to what blocks, methods, class bodies and module bodies do.)
So, what does it mean to return the first object that determines the outcome? That's simple, really: the result of the expression
a && b
is truthy if both a and b are truthy, otherwise it is falsy. So, if a is falsy, it is completely irrelevant what b is: the result will be falsy either way. So, we might just as well simply return a. (Remember, a doesn't have to be false, it could also be nil and the programmer might want to know which one of the two it was.)
If, OTOH, a is truthy (IOW it is neither nil nor false), then the result of the whole expression is solely dependent on b: if b is truthy, the whole result will be truthy, otherwise if b is falsy, the whole result will be falsy. So, we might just as well return b itself, instead of first converting it to a Boolean.
|| and or are analogous or more precisely dual to && and and.
You posted this example:
id = current_user && current_user.record.id
Here, the author isn't even expecting current_user to be a Boolean value! Instead, he expects it to be either a User or nil. But that's perfectly fine, because a User is truthy and nil is falsy, so they still can be used in a Boolean expression.
The basic intention is that the author wants to prevent a NoMethodError exception being raised, if he tries to call #record on nil.
An alternative way of expressing this would be
id = current_user.record.id unless current_user.nil?
If you want all the gory details, check out Section 11.1 (page 36) of the Draft ISO Ruby Specification or the excutable specifications of the RubySpec project. (Here's the one for &&.)
I wrote a pure Ruby implementation of Ruby's Boolean operators and conditional expressions once for fun. The meat of the implementations is these two mixins.
The logical and is short circuiting. That means that if the construct is X && Y, and X is false then Y never gets checked because the whole thing is certainly going to be yield false.
That code is saying, essentially:
if (current_user is TRUE) {
id = current_user.record.id;
]
Here's some console output showing you get the second value if the first is true:
irb(main):005:0> true && 9
=> 9
and nil if the first is nil:
irb(main):008:0> nil && 9
=> nil

What does !! mean in ruby?

Just wondering what !! is in Ruby.
Not not.
It's used to convert a value to a boolean:
!!nil #=> false
!!"abc" #=> true
!!false #=> false
It's usually not necessary to use though since the only false values to Ruby are nil and false, so it's usually best to let that convention stand.
Think of it as
!(!some_val)
One thing that is it used for legitimately is preventing a huge chunk of data from being returned. For example you probably don't want to return 3MB of image data in your has_image? method, or you may not want to return your entire user object in the logged_in? method. Using !! converts these objects to a simple true/false.
It returns true if the object on the right is not nil and not false, false if it is nil or false
def logged_in?
!!#current_user
end
! means negate boolean state, two !s is nothing special, other than a double negation.
!true == false
# => true
It is commonly used to force a method to return a boolean. It will detect any kind of truthiness, such as string, integers and what not, and turn it into a boolean.
!"wtf"
# => false
!!"wtf"
# => true
A more real use case:
def title
"I return a string."
end
def title_exists?
!!title
end
This is useful when you want to make sure that a boolean is returned. IMHO it's kind of pointless, though, seeing that both if 'some string' and if true is the exact same flow, but some people find it useful to explicitly return a boolean.
Note that this idiom exists in other programming languages as well. C didn't have an intrinsic bool type, so all booleans were typed as int instead, with canonical values of 0 or 1. Takes this example (parentheses added for clarity):
!(1234) == 0
!(0) == 1
!(!(1234)) == 1
The "not-not" syntax converts any non-zero integer to 1, the canonical boolean true value.
In general, though, I find it much better to put in a reasonable comparison than to use this uncommon idiom:
int x = 1234;
if (!!x); // wtf mate
if (x != 0); // obvious
It's useful if you need to do an exclusive or. Copying from Matt Van Horn's answer with slight modifications:
1 ^ true
TypeError: can't convert true into Integer
!!1 ^ !!true
=> false
I used it to ensure two variables were either both nil, or both not nil.
raise "Inconsistency" if !!a ^ !!b
It is "double-negative", but the practice is being discouraged. If you're using rubocop, you'll see it complain on such code with a Style/DoubleNegation violation.
The rationale states:
As this is both cryptic and usually redundant, it should be avoided
[then paraphrasing:] Change !!something to !something.nil?
Understanding how it works can be useful if you need to convert, say, an enumeration into a boolean. I have code that does exactly that, using the classy_enum gem:
class LinkStatus < ClassyEnum::Base
def !
return true
end
end
class LinkStatus::No < LinkStatus
end
class LinkStatus::Claimed < LinkStatus
def !
return false
end
end
class LinkStatus::Confirmed < LinkStatus
def !
return false
end
end
class LinkStatus::Denied < LinkStatus
end
Then in service code I have, for example:
raise Application::Error unless !!object.link_status # => raises exception for "No" and "Denied" states.
Effectively the bangbang operator has become what I might otherwise have written as a method called to_bool.
Other answers have discussed what !! does and whether it is good practice or not.
However, none of the answers give the "standard Ruby way" of casting a value into a boolean.
true & variable
TrueClass, the class of the Ruby value true, implements a method &, which is documented as follows:
Returns false if obj is nil or false, true otherwise.
Why use a dirty double-negation when the standard library has you covered?

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