模式匹配的基本问题
我注意到 [H|T] = [1]. 成功，但 [H|T] = []. 失败。我想这就是它的工作原理，但是设计者有什么原因没有选择让这个模式匹配成功并导致分配 H=[] 和 T=[]< /代码>？
9> [H|T] = [1].
[1]
10> H.
1
11> T.
[]
12> [H|T] = [].
** exception error: no match of right hand side value []
I noticed that [H|T] = [1]. succeeds but [H|T] = []. fails. I guess that's just how it works, but is there any reason the designer didn't chose to let this pattern matching succeed and result in assignment of H=[] and T=[]?
9> [H|T] = [1].
[1]
10> H.
1
11> T.
[]
12> [H|T] = [].
** exception error: no match of right hand side value []
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如果
[H|T]将[]与H=T=[]匹配，则[[]] 与[]区分开来。此外，模式
[]和[H|T]将不再是互斥的，因此如果您不小心先匹配了[H|T]在递归函数中，其中[]是基本情况，您会导致无限递归。另外，使用
[]作为“此列表没有头部”的符号似乎很随意，可能会让很多用户感到惊讶。If
[H|T]would match[]withH=T=[], then[[]]would not be distinguishable from[]using pattern matching.Further the patterns
[]and[H|T]would no longer be mutually exclusive, so if you accidentally matched[H|T]first in a recursive function, where[]is the base case, you'd cause infinite recursion.Also using
[]as a symbol for "this list does not have a head" seems quite arbitrary and might surprise a lot of users.虽然 @sepp2k 所说的是正确的，但
[]不匹配[_|_]的更根本原因是它们是不同的数据类型，因此不应该匹配。它会破坏模式匹配的要点。While what @sepp2k says is correct, a more fundamental reason for
[]not to match[_|_]is that they are different data types and so should not match. It would defeat the point of pattern matching.