重新定义中缀操作符(Redefine the infix operators)

 我正在学习Jason Hickey的Objective Caml简介 。 
 
 只是有一个关于重新定义中缀操作符的问题 。 
 
 
 所以在书中有这样一段话： 
 
# let (+) = ( * )
      and (-) = (+)
      and ( * ) = (/)
      and (/) = (-);;
  val + : int > int > int = <fun>
  val - : int > int > int = <fun>
  val * : int > int > int = <fun>
  val / : int > int > int = <fun>
# 5 + 4 / 1;;
-: **int = 15**
 
 
 首先，这些重新定义是如何工作的？ 
 
 对我来说，似乎函数运行在一种无限循环中，因为所有的操作都被重新定义和连接。 
 
 例如，如果我做1+2 ，那么它将是1 * 2并且因为( * ) = (/) ，它将是1 / 2并且因为(/) = (-) ，那么它将是1-2 ，等等等等。 我对吗？ 
 
 
 其次， 5 + 4 / 1的结果是15 ，即使这些函数在重定义中只执行了一步？ 
 
 因此，假定重新定义将进一步执行，即1 + 2将仅为1 * 2并且不再有变换，所以5 + 4 / 1应该是5 * 4 -1 ，对吗？ 那么答案是19 。 我对么？ 

I am learning Jason Hickey's Introduction to Objective Caml.
 
Just have a question about Redefine the infix operators.
 
 
So in the book, there is such a paragraph:
 
# let (+) = ( * )
      and (-) = (+)
      and ( * ) = (/)
      and (/) = (-);;
  val + : int > int > int = <fun>
  val - : int > int > int = <fun>
  val * : int > int > int = <fun>
  val / : int > int > int = <fun>
# 5 + 4 / 1;;
-: **int = 15**
 
 
First, how does these redefinition work?
 
To me, it seems the functions are running in a kind of indefinite loop, because all the operations seem redefined and connected.
 
for example, if I do 1+2, then it will be 1 * 2 and since ( * ) = (/), it will be then 1 / 2 and since (/) = (-), then it will be 1-2, so on so forth. Am I right?
 
 
Second, will the result of 5 + 4 / 1 be 15, even if the functions are executed only one step further in the redefinition?
 
So assume the redefinition will be execute one step further, i.e., 1 + 2 will only be 1 * 2 and no more transform, so 5 + 4 / 1 should be 5 * 4 -1, right? then the answer is 19. Am I correct?

原文：https://stackoverflow.com/questions/13686223