From: Bulat Ziganshin <bulatz@...>
Date: 2002-11-15T15:27:29+09:00
Subject: Re: Syntactic Sugar Question

Hello Bret,

Friday, November 15, 2002, 3:01:26 AM, you wrote:

BJ> foo ||= "oink"
BJ> foo = (foo || "oink")

from the "programming ruby":

Variable/Method Ambiguity

When Ruby sees a name such as ``a'' in an expression, it needs to determine if it is a local variable reference or a call to a method with no parameters. To decide which is the case, Ruby uses a heuristic. As Ruby reads a source file, it keeps track of symbols that have been assigned to. It assumes that these symbols are variables. When it subsequently comes across a symbol that might be either a variable or a method call, it checks to see if it has seen a prior assignment to that symbol. If so, it treats the symbol as a variable; otherwise it treats it as a method call. As a somewhat pathological case of this, consider the following code fragment, submitted by Clemens Hintze. 

def a
  print "Function 'a' called\n"
  99
end

for i in 1..2
  if i == 2
    print "a=", a, "\n"
  else
    a = 1
    print "a=", a, "\n"
  end
end

 
produces: a=1
Function 'a' called
a=99

 


During the parse, Ruby sees the use of ``a'' in the first print statement and, as it hasn't yet seen any assignment to ``a,'' assumes that it is a method call. By the time it gets to the second print statement, though, it has seen an assignment, and so treats ``a'' as a variable. 

Note that the assignment does not have to be executed---Ruby just has to have seen it. This program does not raise an error. 

a = 1 if false; a

 



-- 
Best regards,
 Bulat                            mailto:bulatz@integ.ru