[Haskell-beginners] Continuations

[Joel Neely]

Shishir,

You've already gotten some excellent descriptions.

As a recovering OO programmer, I get to the same answer by a slightly
different path, offered below for what it's worth.

When I read the first line:

let add_cps x y = \k -> k (x+y)


I see a definition of a function (add_cps) with two arguments (x and y)
whose evaluation yields a function (\k -> k (x+y)).

*I started to write "an anonymous function" out of habit, but stopped
myself. I don't think of functions as having names or not having names, any
more than I think of expressions as having names or not. Variables are
bound to (refer to) values, and a function is just another kind of value to
which a variable can refer. (And multiple variables can refer to the same
value, by the way, so none of them is really "the name" of the value,
regardless of whether that value happens to be a function.)*


So at this point, I can think of using the name add_cps in another
expression, such as:

add_cps 3 4


which (by the substitution principle) must be equivalent to:

\k -> k (3+4)


That expression is clearly a function (because of the lambda) that takes a
single argument and applies it to the expression (3+4) (which isn't yet
evaluated, by the way, but when it is I expect it to yield 7). So the
argument to that function expression must itself be a function.

When I get to the second line:

add_cps 3 4 $ print


I confess to cheating a bit (and the experts may want to offer me some
correction). I mentally rewrite that as:

(add_cps 3 4) print


because the precedence rules tell me to evaluate the left argument of $ before
combining that with what follows.

So I can apply the substitution principle to get

(\k -> k (3+4)) print


and then apply it again to get

print (3+4)


Hope this helps,
-jn-


On Thu, May 28, 2015 at 8:57 AM, Shishir Srivastava <
shishir.srivastava at gmail.com> wrote:

> Hi,
>
> Reading on continuation I've came across this new style of creating the
> functions which I guess is not very clear in how it works
>
> ---------------
> Prelude> let add_cps x y = \k -> k (x+y)
> Prelude> add_cps 3 4 $ print
> 7
> ---------------
>
> I have some questions as to
> 1) what is the role of variable 'k' and what eventually happens to it.
> 2) How does print work after the $ because there is clearly no parameter
> being passed to it.
>
> Thanks,
> Shishir Srivastava
>
>
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-- 
Beauty of style and harmony and grace and good rhythm depend on simplicity.
- Plato
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