Re: why is $1 in a grep() equal to nil?

Eric Christopherson wrote in post #983739:
> On Thu, Feb 24, 2011 at 2:59 PM, 7stud -- <[e-mail]>
> wrote:
>>  end
>> ---get_cpu_info---
>>      puts "-->#{$1}<---"
>>        end
>> comp1 = Computer.new(1, DataSource.new)
>> puts comp1.mouse
>>
>>
>> --output:--
>> Line 32:in `+': can't convert nil into String (TypeError)
>> from t.rb:32:in `mouse'
>> from t.rb:52
>
> I'm not sure of the specifics, but $1 doesn't persist outside of the
> block created in the grep statement. When you call comp1.mouse, that's
> no longer within that block -- the method was defined in it, but once
> it was made a method it took on an existence of its own.

$1 is a *global* variable, so saying it doesn't persist outside of a 
block doesn't make any sense.

I think what is happening is that the body of the define_method() call 
forms a closure around the variable $1.  Unfortunately, the problem with 
global variables is that other parts of the code can change their value. 
In this instance, I think a subsequent unsuccessful pattern match 
assigns nil to $1.  Here is an example of that:

arr = ["hello"]

arr.each do |x|
    x =~ /h(.)ll/
    puts $1
end

puts $1

"hello" =~ /xxx/

puts $1

--output:--
e
e
nil


There are a lot other methods in the DataSource class that are inherited 
by all classes, and the last one in the list must not be one of the 
methods I defined, so the pattern match fails against that method name, 
and $1 gets set to nil.  Subsequently, when I call the mouse() method, 
it reads the current value of $1, which is nil.

Assigning $1 to a local variable, like 'name', means that each method 
created by define_method() gets its own distinct 'name' variable in the 
method body.

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