这是因为 [x] 等于 [x | []] ，所以它匹配 f2（[_ | B]） - > [B]; 。你可以看到 B = [] 你的情况。

it is because [x] is equal to [x|[]] so it matches f2([_|B]) -> [B];. As you can see B=[] inn your case.

我想你没有写你想要的去做。在表达式 [A | B] 中，A是列表的第一个元素，而B是列表的其余部分（因此它是一个列表）。这意味着 [1,2,1] 将不匹配 [A1，A2 | A1] ;但是 [[1]，2,1] 或 [[a，b]，1，a，b] 将会。

I think you didn't write what you want to do. in the expression [A|B], A is the first element of the list, while B is the rest of the list (so it is a list). That means that [1,2,1] will not match [A1, A2 | A1]; but [[1],2,1] or [[a,b],1,a,b] will.