Correct terminology for a function that yields

ruby-talk
1

def foo
  yield
end

foo do
  puts "hey"
end

What's the correct description of what is going on there? I've got a
function foo. No arguments, right? Does it "take a block"? I know
it gives control to the block, but I'm not sure of the correct
terminology to use.

Joe

2

It does indeed "take a block". The block argument is implicit. You
could make it explicit if you wanted to:

def foo(&blk)
    blk.call
end

In more general terms, any function that takes another function as an
argument (or returns one as a result) is known as a higher-order
function.

regards,
Ed

···

On Fri, Dec 23, 2005 at 09:00:31AM +0900, Joe Van Dyk wrote:

def foo
  yield
end

foo do
  puts "hey"
end

What's the correct description of what is going on there? I've got a
function foo. No arguments, right? Does it "take a block"? I know
it gives control to the block, but I'm not sure of the correct
terminology to use.

3

Ok, thanks. I'm writing documentation for a domain-specific language
(in Ruby) that I created. People who use it are probably programmers,
but may not be experienced with Ruby, so I wanted to explain a little
bit about what's going on behind the scenes.

···

On 12/22/05, Edward Faulkner <ef@alum.mit.edu> wrote:

On Fri, Dec 23, 2005 at 09:00:31AM +0900, Joe Van Dyk wrote:
> def foo
> yield
> end
>
> foo do
> puts "hey"
> end
>
>
> What's the correct description of what is going on there? I've got a
> function foo. No arguments, right? Does it "take a block"? I know
> it gives control to the block, but I'm not sure of the correct
> terminology to use.

It does indeed "take a block". The block argument is implicit. You
could make it explicit if you wanted to:

def foo(&blk)
    blk.call
end

In more general terms, any function that takes another function as an
argument (or returns one as a result) is known as a higher-order
function.

4

Joe Van Dyk wrote:

def foo
  yield
end

foo do
  puts "hey"
end

What's the correct description of what is going on there? I've got
a function foo. No arguments, right? Does it "take a block"? I
know it gives control to the block, but I'm not sure of the correct
terminology to use.

It does indeed "take a block". The block argument is implicit. You
could make it explicit if you wanted to:

def foo(&blk)
    blk.call
end

Note, that this has some performance implications though.

In more general terms, any function that takes another function as an
argument (or returns one as a result) is known as a higher-order
function.

Ok, thanks. I'm writing documentation for a domain-specific language
(in Ruby) that I created. People who use it are probably programmers,
but may not be experienced with Ruby, so I wanted to explain a little
bit about what's going on behind the scenes.

You can as well call the block "anonymous function" or "anonymous
callback" IMHO. What I like about the "callback" variant is that it
precisely describes what's happening here: the caller provides a function
as hook that is called by the method.

Kind regards

    robert

···

On 12/22/05, Edward Faulkner <ef@alum.mit.edu> wrote:

On Fri, Dec 23, 2005 at 09:00:31AM +0900, Joe Van Dyk wrote: