In order for the CPS to work, we have to adopt a new restriction on
beta reduction: beta reduction does not occur underneath a lambda.
That is, `(\x.y)z` reduces to `z`, but `\u.(\x.y)z` does not reduce to
-`\w.z`, because the `\w` protects the redex in the body from
-reduction. (In this context, a redex is a part of a term that matches
+`\u.z`, because the `\u` protects the redex in the body from
+reduction. (In this context, a "redex" is a part of a term that matches
the pattern `...((\xM)N)...`, i.e., something that can potentially be
the target of beta reduction.)
Start with a simple form that has two different reduction paths:
-reducing the leftmost lambda first: `(\x.y)((\x.z)w) ~~> y`
+reducing the leftmost lambda first: `(\x.y)((\x.z)u) ~~> y`
-reducing the rightmost lambda first: `(\x.y)((\x.z)w) ~~> (\x.y)z ~~> y`
+reducing the rightmost lambda first: `(\x.y)((\x.z)u) ~~> (\x.y)z ~~> y`
After using the following call-by-name CPS transform---and assuming
that we never evaluate redexes protected by a lambda---only the first
[\xM] = \k.k(\x[M])
[MN] = \k.[M](\m.m[N]k)
-Here's the result of applying the transform to our problem term:
+Here's the result of applying the transform to our simple example:
[(\x.y)((\x.z)u)] =
\k.[\x.y](\m.m[(\x.z)u]k) =
[\x.y](\m.m[(\x.z)u] I) =
(\k.k(\x.y))(\m.m[(\x.z)u] I)
* *
- (\x.y)[(\x.z)u] I
+ (\x.y)[(\x.z)u] I --A--
*
y I
The application to `I` unlocks the leftmost functor. Because that
-functor (`\x.y`) throws away its argument, we never need to expand the
-CPS transform of the argument.
+functor (`\x.y`) throws away its argument (consider the reduction in the
+line marked (A)), we never need to expand the
+CPS transform of the argument. This means that we never bother to
+reduce redexes inside the argument.
Compare with a call-by-value xform:
This time the reduction unfolds in a different manner:
- {(\x.y)((\x.z)w)} I =
+ {(\x.y)((\x.z)u)} I =
(\k.{\x.y}(\m.{(\x.z)u}(\n.mnk))) I
*
{\x.y}(\m.{(\x.z)u}(\n.mnI)) =
{u}(\n.(\x.{z})n(\n.(\x.{y})nI)) =
(\k.ku)(\n.(\x.{z})n(\n.(\x.{y})nI))
* *
- (\x.{z})u(\n.(\x.{y})nI)
+ (\x.{z})u(\n.(\x.{y})nI) --A--
*
{z}(\n.(\x.{y})nI) =
(\k.kz)(\n.(\x.{y})nI)
*
I y
+In this case, the argument does get evaluated: consider the reduction
+in the line marked (A).
+
Both xforms make the following guarantee: as long as redexes
underneath a lambda are never evaluated, there will be at most one
reduction available at any step in the evaluation.