sicp-ex-1.5


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, to 0:

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

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