严格语言中let语句内(self self)调用及Y-combinator代码推导咨询

Great question—this is a critical detail that trips up many people when working through Y-combinator derivations, especially when moving from lazy to strict evaluation semantics. Let’s break this down step by step.

First, let’s recap the two versions of your code to set the context:

Original Working Version (Works in Both Lazy and Strict Languages) #

(define (part-factorial self)
  (lambda (n)
    (if (= n 0)
        1
        (* n ((self self) (- n 1))))))

((part-factorial part-factorial) 5) ; => 120
(define factorial (part-factorial part-factorial))
(factorial 5) ; => 120

In this version, (self self) is nested inside the lambda that takes n. In strict languages, this means (self self) only gets evaluated when we actually call that lambda (i.e., when we pass a value for n). The first call to (part-factorial part-factorial) returns the lambda immediately—no recursive evaluation happens yet. It’s only when we invoke that lambda (with 5, for example) that (self self) is evaluated to get the recursive factorial function.

The Modified Version (Problematic in Strict Languages) #

(define (part-factorial self)
  (let ((f (self self))) ; Here's the trouble spot!
    (lambda (n)
      (if (= n 0)
          1
          (* n (f (- n 1)))))))

(define factorial (part-factorial part-factorial))

The core issue here is how strict evaluation languages handle let bindings: they evaluate the right-hand side of the binding immediately, before proceeding to the body of the let.

When you run (part-factorial part-factorial), the interpreter first tries to compute f = (self self)—but self is part-factorial, so this is equivalent to calling (part-factorial part-factorial) again. This triggers an infinite recursion loop: each call to part-factorial tries to evaluate (self self) before it can return the lambda, leading to a stack overflow before we ever get to use the factorial function.

Fixing It for Strict Languages #

To make this work in strict languages, we need to delay the evaluation of (self self) until it’s actually needed. We do this by wrapping (self self) in a lambda, so f becomes a function that, when called, evaluates (self self):

(define (part-factorial self)
  (let ((f (lambda (x) ((self self) x)))) ; Delay evaluation!
    (lambda (n)
      (if (= n 0)
          1
          (* n (f (- n 1)))))))

Now, when we call (part-factorial part-factorial), the let binding creates f as a lambda (not an evaluated recursive call). The outer lambda is returned immediately. When we invoke f inside the factorial logic, that’s when (self self) gets evaluated—breaking the infinite recursion and allowing the recursive steps to proceed normally.

This adjustment is exactly what defines the "strict Y-combinator" (sometimes called the Z-combinator) because it ensures self-reference doesn’t trigger an infinite loop before the function is ready to be used.

内容的提问来源于stack exchange，提问作者user8554766