Ruby variable assignment in a conditional "if" modifier
if-statement, local-variables, ruby, variable-assignment
Solution
Read it carefully :
Another commonly confusing case is when using a modifier if:
p a if a = 0.zero?
Rather than printing `true` you receive a NameError, “undefined local variable or method 'a'”. Since Ruby parses the bare `a` left of the `if` first and has not yet seen an assignment to a it assumes you wish to call a method. Ruby then sees the assignment to `a` and will assume you are referencing a `local method`.
The confusion comes from the out-of-order execution of the expression. First the local variable is assigned-to then you attempt to call a nonexistent method.
As you said - None `return foo if (foo = bar.some_method)` and `return foo if (true && (foo = bar.some_method))` will work, I bet you, it wouldn't work, if you didn't define `foo` before this line.
Problem
I have a question about how the Ruby interpreter assigns variables: I use this quite often: ``` return foo if (foo = bar.some_method) ``` where some_method returns an object or nil. However, when I try this: ``` return foo if (true && (foo = bar.some_method)) ``` I get: NameError: undefined local variable or method foo for main:Object. What is the difference in evaluation between the first and second lines that causes the second line to error?