OCaml中的模式匹配性能(Pattern Matching performance in OCaml)

OCaml中的模式匹配性能(Pattern Matching performance in OCaml)

假设我有一个类型type 'a tree = Node of int * 'a * 'a tree

int部分是rank 。

我还有一个函数let rank = function Node (r, _, _) -> r 。

假设在我的实际代码中，我有很多地方需要访问节点的rank ，并且在很多时候，相同节点的rank 。

我的问题是，我是否应该直接使用rank t函数，或者我应该将r的等级赋予r ，然后使用r ？

例如，我可以做

if rank t1 < rank t2 then Node (rank t1 + 1, 5, t1) else Node (rank t2 + 1, 5, t2)

要么

let r1 = rank t1 in let r2 = rank t2 in if r1 < r2 then Node (r1+1, 5, t1) else Node (r2, 5, t2)

什么是性能差异？ 哪种方式更好，为什么？

Let's say I have such a type type 'a tree = Node of int * 'a * 'a tree

The int part is the rank.

Also I have a function let rank = function Node (r, _, _) -> r.

Suppose in my real code, I have lots of places that need to access the rank of a node, and much of the time, the rank of same nodes.

My question is that Should I always use rank t function directly or should I give the rank of a t to, say, r, then use r?

For example, I can do

if rank t1 < rank t2 then Node (rank t1 + 1, 5, t1) else Node (rank t2 + 1, 5, t2)

or

let r1 = rank t1 in let r2 = rank t2 in if r1 < r2 then Node (r1+1, 5, t1) else Node (r2, 5, t2)

What is the performance diff? And which way is better and why?

最满意答案

我认为使用let来避免计算两次排名更好。 我不认为OCaml编译器优化两次调用相同的函数; 它需要某种效果分析。

我会很高兴错了;）

I think using let to avoid computing the rank twice is better. I don't think the OCaml compiler optimizes calling the same function twice; it would require some kind of effect analysis.

I'd be glad to be wrong though ;)