Iterators and Assigment

[Godspeed]

Consider...

def fred
yield $k
puts "in fred #{$k}"
end

$k=2
fred {|o| o = 3}
puts "outside fred #{$k}"

produces:

in fred 2
outside fred 2

why is $k still 2 after the call to fred?

Thanks in advance.

 

[Robert Klemme]

Consider...

def fred
yield $k
puts "in fred #{$k}"
end

$k=2
fred {|o| o = 3}
puts "outside fred #{$k}"

produces:

in fred 2
outside fred 2

why is $k still 2 after the call to fred?

Because Ruby does call by value where values are object references. Same
holds for blocks and their parameters. There is no implicit aliasing
between k$ and o. When the block is invoked you get a new reference which
initially points to the same instance as $k. Then you assign it and it
points to some other object leaving $k still pointing to the old instance.

Kind regards

robert

 

[Jonathan Leighton]

Because Ruby does call by value where values are object references. Same
holds for blocks and their parameters. There is no implicit aliasing
between k$ and o. When the block is invoked you get a new reference which
initially points to the same instance as $k. Then you assign it and it
points to some other object leaving $k still pointing to the old instance.

Kind regards

robert

So, try this:

def fred
$k = yield
puts "in fred #{$k}"
end

$k = 2
fred { 3 }
puts "outside fred #{$k}"

Returns:

in fred 3
outside fred 3

You can call yield with parameters if you need the value of $k to be
accessible inside the block. Although $k is accessible inside the block
anyway as it's a global. Although you're probably not using a global in
your actual code.

Regards

 

[Godspeed]

Thanks guys. Your responses have cleared this up. I was assumig some sort of
substitution (of the block into the def) going on, but I realise it is the
other way round, the parameters to yield are parameters to the block. Makes
perfect sense now.