I was expecting the following snippet:
var = "Not Empty" unless defined? var
var # => nil
to return "Not Empty", but I got nil. Any insight into why this is happening?
This is one of the only moments in Ruby I would call actual WTFs.
You have to use
unless defined? var
var = :value
end
With the postfix syntax, the interpreter will internally nil-ify the value so it can reason about the variable, thus making it defined before the check is done:
# Doesn't print anything
unless defined?(foo) and (p(foo) or true)
foo = :value
end
# Prints nil
bar = :value unless defined?(bar) and (p(bar) or true)
Local variables are defined (as nil) at the point they are parsed. Definition of var2 precedes the condition. That makes var2 defined even when if the assignment is not executed. Then, the condition evaluates that var2 is defined, which retains the value nil for var2.
Related
Today I came across an interesting piece of code. It's more like a scientific question about the ruby parser.
We know everything in ruby is an object and every expression evaluates at least to nil.
But how is the following "assignment" parsed:
somevar
NameError: undefined local variable or method 'somevar' for main:Object
somevar = "test" if false
=> nil
somevar
=> nil
You see the variable is undefined until it's used in the assignment. But the assignment is not happening because of the condition. Or is it happening because the condition evaluates to nil? I tried something which would break in this case, but it just works:
a = {}
a[1/0]
ZeroDivisionError: divided by 0
a[1/0] = "test" if false
=> nil
So is this meant to work the way it is? Or does it make sense to test the variable (defined?(somevar)) before accessing, in case a future version of ruby will break this behaviour? As example by saving the assigned pointer to this variable.
My currently used ruby version is 3.0.2.
This is expected behavior in Ruby. Quote from the Ruby docs:
The local variable is created when the parser encounters the assignment, not when the assignment occurs:
a = 0 if false # does not assign to a
p local_variables # prints [:a]
p a # prints nil
If you do = "test" if false it evaluates to nil => no assignment needed. But by calling somevar = ... you told the interpreter to declare the name somevar. The nil aren't the same (if that makes sense).
The [] operator however doesn't declare a variable (only accesses) but since if false isn't true there is no assignment so the whole left side isnt evaluated.
Consider:
a = [1,2,3]
a[1] = "test" if false
a
=> [1,2,3]
a[1] is neither nil nor test.
Not sure what you expect or how future Ruby will break this?
I have a curiosity with variable definition. I know about variable definition, it's just to understand how it works. The idea is to set myvar to 0 if it is not defined. I don't want to use false because false could be returned by false or nil.
def foo
myvar=1000
myvar = ((defined? myvar)==nil) ? 0 : myvar
puts myvar
end
foo # => 1000
but
def foo
# myvar=1000
myvar = ((defined? myvar)==nil) ? 0 : myvar
puts myvar
end
foo # => nil
but
def foo
mybool = ((defined? myvar)==nil)
myvar = mybool ? 0 : myvar
puts myvar
end
foo # => 0 it rocks !
It's like if the boolean test was affected to myvar before the final test was evaluated.
((defined? myvar)==nil) gives 2 possibilities defined? myvar could be nil or the right side of test nil could be nil (it's nil).
To get rid of the nil part of this test, I tried this code :
def foo
mylocalvariable = 2 # it gives 'local-variable' and not nil
puts ((defined? myvar) == (defined? mylocalvariable ))
myvar = ((defined? myvar)!=(defined? mylocalvariable )) ? 0 : myvar
puts myvar
end
foo # => nil
It seems the first part of the test was affected to myvar, but assignment operators come after () or after comparison operator. Do you have an idea? I don't want an other code but just why it works like that.
Thank you.
You might not want other code, but you need some because you're going the long way around to accomplish something that is simple. If you want to set an undefined variable to 0 use:
myvar ||= 0
How does it work? ||= is a conditional assignment operator, that basically looks at the variable being assigned to on the left, and, if it's nil or false, assigns the value on the right to it. It's a trick found in many languages.
Ternary (?:) statements are useful for small if/then/else statements, but aren't really the right choice for what you want to do. Instead, if you want to use something besides ||=, use:
myvar = 0 unless myvar
Again, all of these will reassign myvar if it happens to be a false value, not just if it's nil. That's a potential problem, but, as developers, we're supposed to be in control and should have a good idea what type of value a variable is holding at any moment so it shouldn't be that big of a problem.
Hi this kind of operator is ternary operator.
Example:
#x = 1
#y = 2
#x>2 ? puts "YES" : puts "NO"
You can also check if it for an ActiveRecord
Example:
#x = User.find(:all, :conditions => ["id = ?", 1])
#x.nil? ? puts "Null" : puts "Not Null"
I know why. When you code myvar= and you test defined?myvar, ruby consider myvar defined. If you remove myvar= , myvar is considered undefined in the test.
The left part is read before the right part.
I could not find anywhere a formal specification of the if (or unless) modifier:
123 if x > 0
what is the value of the above statement if x is not above zero? irb suggests nil, but is it documented anywhere?
(yes, this is perhaps a stupid question, sorry, could not find a spec).
An expression that is not evaluated is the same as not existing. And your condition should be part of some code block such as definition or the main environment, etc. A code block without a content is evaluated to nil.
class A; end
# => nil
def foo; end
foo
# => nil
()
# => nil
begin; end
# => nil
eval("")
# => nil
So the reason your example returns nil has nothing to do with condition itself. It is just due to there being no evaluation.
Yes, it is nil. What else could it return? The statement in the if clause is used for the if itself, and the then statement is not executed. There is nothing to return, so nil.
Relevant spec
it "returns nil if else-body is empty and expression is false" do
if false
123
else
end.should == nil
end
What's going on here? What is the subtle difference between the two forms of "unless"?
> irb(main):001:0> foo = true unless defined?(foo)
=> nil
irb(main):002:0> unless defined?(fooo) ; fooo = false ; end
=> false
thx
Apparently, ruby creates local variable at parse time setting them to nilso it is defined and this is done whether the code is executed or not.
When the code is evaluated at your first line, it doesn't execute the assignment part since foo is set to nil. In the second line, because fooo has not been parsed yet, defined?returns nil letting the code inside the block execute and assign fooo.
As an example you can try this:
if false
foo = 43
end
defined? foo
=> "local-variable"
This is taken from a forum post at ruby-forum.
Let's start with something simpler:
# z is not yet defined
irb(main):001:0> defined?(z)
=> nil
# Even though the assignment won't execute,
# the mere presence of the assignment statement
# causes z to come to life.
irb(main):002:0> z = 123 if false
=> nil
irb(main):003:0> defined?(z)
=> "local-variable"
irb(main):004:0> z
=> nil
Now we can figure out your first example.
foo = true unless defined?(foo)
Is foo defined? Before we press ENTER in irb, no. However, the presence of the assignment statement causes foo to come to life. That means the assignment statement won't be executed, leaving foo in existence but having nil as its value. And what is the last expression evaluated in the irb line? It is unless defined?(foo), which evaluates to nil.
For more info on how assignments (even those that do not get executed) cause variables to exist, see this discussion of Variable/Method Ambiguity.
In your second example, there is nothing mysterious at all: fooo is not defined, so the code in the block executes, setting fooo to false. That assignment is the last expression evaluated, so false is the return value of our block.
irb(main)> foo = true unless defined?(Integer)
=> nil
irb(main)> foo = true unless defined?(thisIsUndefined)
=> true
Your first block is returning nil because the way it's written leaves 2 options:
foo is not defined --> assign true
foo is defined --> do nothing
Here, foo must be defined when the line is evaluated. Thus, nothing happens and nil is returned.
irb(main)> unless defined?(Integer) ; fooo = false ; end
=> nil
irb(main)> unless defined?(thisIsUndefined) ; fooo = false ; end
=> false
Your second block operates the same way your first one does. If fooo is not defined, the block is entered and fooo is set to false. The result of the last line of the block is the return value of the block, thus the false you are seeing. If fooo does exist, then the block is skipped over and nothing happens, therefore there is nothing to return, therefore the nil.
Based on your code, I would say that foo was defined when this code was run and fooo was not (test code shown was generated in Ruby 1.8.6). If you did not define either of these before running this code, then you may have something called foo that is defined by default (do defined?(foo) by itself to check). Try using a different name and see if you get the same results.
Edit:
irb(main)> defined?(bar)
=> nil
irb(main)> bar = true unless defined?(bar)
=> nil
irb(main)> defined?(bar)
=> "local-variable"
Apparently, defined?() is returning true since it has already seen bar (at the beginning of the line), even though you are still in the process of defining it.
in the first instance you call foo into existence in the assignment statement. Maybe this will clarify:
bar = if true
puts bar.class
else
puts "not reached"
end
NilClass
=> nil
baz = if true
puts baz.class
42
else
puts "not reached"
end
NilClass
=> 42
August, all look fine in 1.8.7:
$ irb
irb(main):001:0> unless defined?(fooo); fooo = true; end
=> true
irb(main):002:0> fooo
=> true
irb(main):003:0> `ruby --version`
=> "ruby 1.8.7 (2008-06-20 patchlevel 22) [i486-linux]\n"
Well.. One form is a block and one form isn't. The second part, the block, returns the last statement evaluated. The first one.. Hrm.. I don't know exactly what it's doing.
In Ruby 1.8.7:
foo = true unless defined?(foo)
p foo # => nil
unless defined?(fooo); fooo = true; end
p foo # => nil
I don't have an explanation for the behaviour you are seeing.
How can I check whether a variable is defined in Ruby? Is there an isset-type method available?
Use the defined? keyword (documentation). It will return a String with the kind of the item, or nil if it doesn’t exist.
>> a = 1
=> 1
>> defined? a
=> "local-variable"
>> defined? b
=> nil
>> defined? nil
=> "nil"
>> defined? String
=> "constant"
>> defined? 1
=> "expression"
As skalee commented: "It is worth noting that variable which is set to nil is initialized."
>> n = nil
>> defined? n
=> "local-variable"
This is useful if you want to do nothing if it does exist but create it if it doesn't exist.
def get_var
#var ||= SomeClass.new()
end
This only creates the new instance once. After that it just keeps returning the var.
The correct syntax for the above statement is:
if (defined?(var)).nil? # will now return true or false
print "var is not defined\n".color(:red)
else
print "var is defined\n".color(:green)
end
substituting (var) with your variable. This syntax will return a true/false value for evaluation in the if statement.
defined?(your_var) will work. Depending on what you're doing you can also do something like your_var.nil?
Try "unless" instead of "if"
a = "apple"
# Note that b is not declared
c = nil
unless defined? a
puts "a is not defined"
end
unless defined? b
puts "b is not defined"
end
unless defined? c
puts "c is not defined"
end
WARNING Re: A Common Ruby Pattern
the defined? method is the answer. See the accepted answer above.
But watch out... consider this common ruby pattern:
def method1
#x ||= method2
end
def method2
nil
end
method2 always returns nil.
The first time you call method1, the #x variable is not set - therefore method2 will be run. and
method2 will set #x to nil.
But what happens the second time you call method1?
Remember #x has already been set to nil. But method2 will still be run again!! If method2 is a costly undertaking this might not be something that you want.
Let the defined? method come to the rescue:
def method1
return #x if defined? #x
#x = method2
end
As with most things, the devil is in the implementation details.
Use defined? YourVariable
Keep it simple silly .. ;)
Here is some code, nothing rocket science but it works well enough
require 'rubygems'
require 'rainbow'
if defined?(var).nil? # .nil? is optional but might make for clearer intent.
print "var is not defined\n".color(:red)
else
print "car is defined\n".color(:green)
end
Clearly, the colouring code is not necessary, just a nice visualation in this toy example.
You can try:
unless defined?(var)
#ruby code goes here
end
=> true
Because it returns a boolean.
As many other examples show you don't actually need a boolean from a method to make logical choices in ruby. It would be a poor form to coerce everything to a boolean unless you actually need a boolean.
But if you absolutely need a boolean. Use !! (bang bang) or "falsy falsy reveals the truth".
› irb
>> a = nil
=> nil
>> defined?(a)
=> "local-variable"
>> defined?(b)
=> nil
>> !!defined?(a)
=> true
>> !!defined?(b)
=> false
Why it doesn't usually pay to coerce:
>> (!!defined?(a) ? "var is defined".colorize(:green) : "var is not defined".colorize(:red)) == (defined?(a) ? "var is defined".colorize(:green) : "var is not defined".colorize(:red))
=> true
Here's an example where it matters because it relies on the implicit coercion of the boolean value to its string representation.
>> puts "var is defined? #{!!defined?(a)} vs #{defined?(a)}"
var is defined? true vs local-variable
=> nil
It should be mentioned that using defined to check if a specific field is set in a hash might behave unexpected:
var = {}
if defined? var['unknown']
puts 'this is unexpected'
end
# will output "this is unexpected"
The syntax is correct here, but defined? var['unknown'] will be evaluated to the string "method", so the if block will be executed
edit: The correct notation for checking if a key exists in a hash would be:
if var.key?('unknown')
Please note the distinction between "defined" and "assigned".
$ ruby -e 'def f; if 1>2; x=99; end;p x, defined? x; end;f'
nil
"local-variable"
x is defined even though it is never assigned!
defined? is great, but if you are in a Rails environment you can also use try, especially in cases where you want to check a dynamic variable name:
foo = 1
my_foo = "foo"
my_bar = "bar"
try(:foo) # => 1
try(:bar) # => nil
try(my_foo) # => 1
try(my_bar) # => nil
Also, you can check if it's defined while in a string via interpolation, if you code:
puts "Is array1 defined and what type is it? #{defined?(#array1)}"
The system will tell you the type if it is defined.
If it is not defined it will just return a warning saying the variable is not initialized.
Hope this helps! :)
Leaving an incredibly simple example in case it helps.
When variable doesn't exist:
if defined? a then "hi" end
# => nil
When variable does exist:
a = 2
if defined? a then "hi" end
# => "hi"