SICP 习题答案1.5

正则序:   完全展开后归约

应用序:   先求值参数而后应用

Using applicative-order evaluation, the evaluation of (test 0 (p)) never terminates, because(p) is infinitely expanded to itself:

 (test 0 (p)) 
  
 (test 0 (p)) 
  
 (test 0 (p)) 

... and so on.

Using normal-order evaluation, the expression evaluates, step by step, to0:

 (test 0 (p)) 
  
 (if (= 0 0) 0 (p)) 
  
 (if #t 0 (p)) 
  
 0