David said:
Hash#each returns an array(as my examples showed). You said:
If you were going to write Hash#each in Ruby, without recourse to
#each,
So, yes...your example would have to return an array to implement
Hash#each.
If you do this:
some_method(some_proc_object)
you're just passing the Proc object around the same way you would pass
a string or an array or any other object.
If you do this:
some_method &some_proc_object
you're telling some_method that you want some_proc_object to play the
special role of code-block.
Ah. So, it's like the splat operator(*)? With the splat operator you
might write:
def meth1(*args) #gathers the args into an array
meth2(*args) #applied a second time -- explodes the array
end
def meth2(a, b)
p a
p b
end
meth1(1, 2)
So, applying & a second time conceptually converts the Proc object back
to a block and passes it to the method?
There could even be a situation where you would do:
some_method(proc_1, proc_2) &proc_3
i.e., send two procs as regular arguments, and use a third one as the
code block.
Whoa. That doesn't seem consistent with your earlier example. Why is
proc3 outside the parentheses? It doesn't seem to work:
def some_method(proc_1, proc_2)
proc_1.call
proc_2.call
yield 'goodbye'
end
p1 = Proc.new{puts "hello"}
p2 = Proc.new{puts "world"}
p3 = Proc.new{|val| puts val}
some_method(p1, p2) &p3
--output:--
hello
world
r6test.rb:5:in `some_method': no block given (LocalJumpError)
from r6test.rb:11
While this works:
p1 = Proc.new{puts "hello"}
p2 = Proc.new{puts "world"}
p3 = Proc.new{|val| puts val}
some_method(p1, p2, &p3)
--output:--
hello
world
goodbye
