s = Proc.new {|x|x*2}
def one_arg(x)
puts yield(x)
end
one_arg(5, &s)
How does one_arg know about &s?
s = Proc.new {|x|x*2} def one_arg(x) puts yield(x) end one_arg(5, &s) How does one_arg
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s = Proc.new {|x|x*2}
def one_arg(x)
puts yield(x)
end
one_arg(5, &s)
How does one_arg know about &s?
The
&operator turns the Proc into a block, so it becomes a one-argument method with a block (which is called withyield). If you had left off the&so that it passed the Proc directly, you would have gotten an error.