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Why in Ruby, a || 1 will throw an error when `a` is undefined, but a = a || 1 will not?

When a is undefined, then a || 1 will throw an error, but a = a || 1 will not. Isn't that a little bit inconsistent?

irb(main):001:0> a
NameError: undefined local variable or method 'a' for main:Object
        from (irb):1
        from c:/ruby/bin/irb:12:in '<main>'

irb(main):002:0> a || 1
NameError: undefined local variable or method 'a' for main:Object
        from (irb):2
        from c:/ruby/bin/irb:12:in '<main>'

irb(main):003:0> a = a || 1
=> 1
like image 957
nonopolarity Avatar asked Jun 06 '10 06:06

nonopolarity


1 Answers

a

Here, you are evaluating a, which isn't defined. Therefore, you get an exception.

a || 1

Here, you still have to evaluate a to determine the value of the boolean expression. Just like above, a isn't defined. Therefore, you get an exception.

a = a || 1

Here, a is defined. It is defined to be an uninitialized local variable. In Ruby, uninitialized variables evaluate to nil, so the right hand side of the assignment expression evaluates to nil || 1 which evaluates to 1, so the return value of the assignment expression is 1 and the side effect is that a is initialized to 1.

EDIT: It seems that there is some confusion on when variables get defined and when they get initialized in Ruby. The get defined at parse time but initialized at runtime. You can see it here:

 foo # => NameError: undefined local variable or method `foo' for main:Object

foo is undefined.

if false
  foo = 'This will never get executed'
end

At this point, foo is defined, even though the line will never get executed. The fact that the line never gets executed is completely irrelevant, because the interpreter has nothing to do with this anyway: local variables are defined by the parser, and the parser obviously sees this line.

foo # => nil

There is no error, because foo is defined, and it evaluates to nil because it is uninitialized.

like image 92
Jörg W Mittag Avatar answered Nov 03 '22 00:11

Jörg W Mittag