From: Justin Johnson <justinj@...>
Date: 2002-07-24T23:47:58+09:00
Subject: Re: Eval question


This is exactly the information I was looking for!

So even this works:

eval "val = 12"
val = val
p val

because at compile time the 'val' variable has been seen now.

I was just making sure that locals are only seen at compile-time, not
run-time and that there was no way of creating them at run-time (eval
compiles code).

--
Justin Johnson.

"ts" <decoux@moulon.inra.fr> wrote in message
news:200207241300.g6OD0wP13014@moulon.inra.fr...
> >>>>> "T" == =?iso-8859-1?Q?Thomas S=F8ndergaard?= <iso-8859-1> writes:
>
> T> So val is created in a scope local to the eval invocation, unless it
> T> already exists in the nesting scope. I think :-)
>
>  No, not really :-)
>
>  For example
>
> pigeon% ruby -e 'eval "a = 12"; eval "p a"'
> 12
> pigeon%
>
>  If the variable `a' is created in a scope local to the first #eval, then
>  this variable will no be accessible for the second #eval
>
> T> puts val # -> 10
>
>  The problem is on this line. This is at *compile* time that ruby make the
>  difference between an access to a local variable or a method call,
>  i.e. ruby must make the choice between
>
>   (1) :  puts val   # I try to access the local variable `val'
>   (2) :  puts val() # I call the method #val
>
>  this choice is made at compile time when ruby know *only* if it exist a
>  local variable and the rule is simple
>
>   * if a variable 'val' was previously found then it make the choice (1)
>   * otherwise it make the choice (2)
>
>  For example with
>
>    eval "val = 12"
>    p val
>
>  when it compile the second line, it has not yet executed the string for
>  #eval this mean that at this step it don't exist a local variable `val'
>  and ruby consider that it's a method call, i.e.
>
>    eval "val = 12"
>    p val()
>
>  At runtime
>    * #eval will create a local variable `val'
>    * *but* in the second line, ruby will try to call the method #val
>
>  Now when you write
>
>    val = nil
>    eval "val = 12"
>    p val
>
>  when it compile the first line, it has found a new variable 'val'
>  when it compile the third line, it know that it exist a variable with the
>  name `val' and in this case it will create the node to access a local
>  variable.
>
>  At runtime, the 3 lines work with the same variable
>
>
> p.s.: You'll have a similar problem with attribute writer
>
>
> Guy Decoux
>