SICP:2,4 序对的过程性表示方法

 

Here are the fully expand steps:

0) at the beginning , we get expression ``(car (cons 1 2))``

1) replace ``cons`` using ``(lambda (m) (m x y))`` , we get ``(car (lambda (m) (m x y))`` .

2) replace ``car`` using ``(lambda (z) (z (lambda (p q) p)``, we get :

((lambda (z) (z (lambda (p q) p))
__(lambda (m) (m 1 2)))

3) replace ``z`` using ``(lambda (m) (m 1 2))`` , we get :

((lambda (m) (m 1 2))
__(lambda (p q) p))

4) replace ``m`` using ``(lambda (p q) p)`` , we get :

((lambda (p q) p)
__1 2)

5) replace ``p`` and ``q`` using ``1`` and ``2, we get :

(lambda (1 2) 1)

6) finally, the answer of the expression is ``1``