I wonder, why is a visible?
if true
puts 'true'
else
puts 'false'
a = 123
end
puts a # no error
# or
# my_hash = {key: a}
# puts my_hash # :key => nil
But this causes an error, even though there will be 'true' shown
if true
puts 'true'
else
puts 'false'
a = 123
end
puts a2 # boooooom
Referencing a inside the if has the effect of declaring it as a variable if there is no method a= defined for the object.
Since Ruby does not require methods to be called using the same syntax as referencing a variable or assigning to one, it needs to make an assessment as to the nature of the token in question. If it could be a method call because a method with that name has been defined, then it will be interpreted as such. If no such method exists at the time the source is compiled, then it will be a variable by default.
if block"? I'm not sure what that means? Variables are scoped, and a is not scoped to the if.
if has no block, only clauses that are conditionally evaluated. The clauses of an if statement do not create a new scope. This is not C.
if as if it was scoped, but you're right, that doesn't appear to be the case. C++ habits refuse to go away. Variables will be scoped to the method they're declared inside of, or the main context if in irb.
a= defined is completely irrelevant. a = foo will always assign to a local variable called a. In fact, that is how Ruby distinguishes between variables and methods in the first place! If you want to call a setter called a=, you have to use self.a = foo. That's why setters are exempt from the privacy rule that private methods can only be called without an explicit receiver, because there wouldn't be a way to call a private setter otherwise.