问题
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?
回答1:
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.
回答2:
You could do return foo||=nil if foo = bar.some_method.
来源：https://stackoverflow.com/questions/21740291/ruby-variable-assignment-in-a-conditional-if-modifier