From: Doug Edmunds <dae_alt3@...>
Date: 2001-03-31T05:40:12+09:00
Subject: [ruby-talk:13363] f.call (x) revisited

Back in Oct 2000, Davide Marchignoli inquired:

>Why the two following (admittedly contrived) computations give
>different results ?

> a = 1 ; x = 1 ; f = proc {|a| a += 1}
> f.call(x)
> f.call(a)
> a # results in 2 <<<???? SEE NOTE BELOW

> a = 1 ; x = 1 ; f = proc {|a| a += 1}
> f.call(a)
> f.call(x)
> a # results in 3 <<<??? SEE NOTE BELOW

I decided to try out his example.  I didn't get his results.
I got 3 for the first result and 2 for the second.
I have expanded the examples to try to see what is going on.
I leave it to the group to explain:
a) why f.call(x) is always 2
b) why f.call(a) is always 3 after f.call(x)
c) where the value of a is coming from in proc {|a| a += 1}

(long code string follows, please don't quote it all!)

-- doug edmunds
30 March 2000

#------code start -------

p "Test 1, where x is called before a"
a = 1; x = 1; f = proc {|a| a += 1}
p "initial value of a:"
p a   #1

p "call x, then a:"
f.call(x)
p a
f.call(a)
p a

p "call x, then a:"
f.call(x)
p a
f.call(a)
p a

p "call x, then a:"
f.call(x)
p a
f.call(a)
p a

p "Test 2, where a is called before x"
a = 1; x = 1; f = proc {|a| a += 1}

p "initial value"
p a  #1 again
p "call a first -- value is 2, only this one time"
f.call(a)
p a  # <<<<< 2, do you see why?

p "call x, then a:"
f.call(x)
p a
f.call(a)
p a

p "call x, then a:"
f.call(x)
p a
f.call(a)
p a

p "call x, then a:"
f.call(x)
p a
f.call(a)
p a

p "Test 3, repeated calls to x"
f.call(x)
p a
f.call(x)
p a
f.call(x)
p a
f.call(x)
p a

p "Test 4, repeated calls to a"
f.call(a)
p a
f.call(a)
p a
f.call(a)
p a
f.call(a)
p a

p "Test 5, back to alternating calls"
p "call x, then a:"
f.call(x)
p a
f.call(a)
p a

# ------- code end --------