如何将quot正确定义为单个函数？无需额外计数器参数的实现方法

Great question! The two-function pattern you’re using right now (a helper with a counter plus a wrapper) is totally valid, but we can clean this up by encapsulating the helper inside the main function so you don’t need separate top-level definitions.

In Haskell, you can use a where clause to define an internal helper function that carries the counter state, while keeping the main function’s interface clean (only taking x and y). Here’s the implementation:

quot x y = go x 0
  where
    go current count
      | current < y = count
      | otherwise = go (current - y) (count + 1)

How this works: #

• The main quot function kicks off the recursion by calling the internal go helper, passing the original x as current and initializing the counter count to 0.
• The go function handles the recursive logic: if current is smaller than y, it returns the accumulated count. Otherwise, it subtracts y from current, increments the count, and recurses.
• Since go is defined inside quot’s where clause, it’s not visible outside the function—so your public interface stays exactly what you want: quot x y with no extra parameters.

This is a standard idiom in functional programming for handling recursive state without cluttering the main function’s signature. You can reuse this pattern for any recursive task where you need to track accumulated values (like counters, sums, or intermediate results) without exposing that state to the caller.

内容的提问来源于stack exchange，提问作者user5775230