CODEWITHSUNDEEP
ruby
uzo
uzo

Reputation: 2821

How are arguments passed to procs?

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?

Upvotes: 0

Views: 95

Answers (2)

glenn jackman
glenn jackman

Reputation: 246847

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

Upvotes: 4

sepp2k
sepp2k

Reputation: 370172

Because the block outputs x*3 (as opposed to s which returns x*2) and 5*3 is 15.

Upvotes: 4

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