s = Proc.new {|x|x*2} puts "proc:" + (s.call(5)).to_s def foo(&a) a.call(5) end foo{|x| puts "foo:" + (x*3).to_s} Running this program produces the output:
proc:10 foo:15 How does the value 3 from the foo block get passed to the proc? I expected this output:
proc:10 foo:10 The proc is always called with the value 5 as the argument because foo is defined as:
a.call(5) Why is foo 15 in the output?
The value 3 does not get passed to the proc because you're not passing s to foo. You probably meant to write
foo {|x| puts "foo: #{s.call(x)}"} or
puts "foo: #{foo(&s)}" Additionally, these are equivalent:
def foo_1(x, &a) puts a.call(x) end def foo_2(x) puts yield(x) end foo_1(5, &s) #=> 10 foo_2(5, &s) #=> 10 Because the block outputs x*3 (as opposed to s which returns x*2) and 5*3 is 15.
a is the block {|x| puts "foo:" + (x*3).to_s}. This block multiplies the argument times 3. So when you call it with the argument 5, you get 5*3, which is 15.