Haskell中实现递归crosssum函数遇线程终止问题求助

First off, that "thread killed" error almost always means your recursion is infinite—the function never hits a base case to stop calling itself. Let's break down how to fix this properly, sticking strictly to the allowed functions.

The Correct Recursive Implementation #

Here's the working version of crosssum that follows all your constraints:

crosssum :: Int -> Int
crosssum n
  | n <= 9    = n  -- Base case: single-digit number is its own crosssum
  | otherwise = (n mod 10) + crosssum (n div 10)

Let's Walk Through Why This Works #

1. Base Case: When n is a single-digit number (n <= 9), we just return n directly. This stops the recursion from going on forever.
2. Recursive Step:
   • n mod 10 grabs the last digit of the number (e.g., 12 mod 10 = 2)
   • n div 10 removes the last digit (e.g., 12 div 10 = 1)
   • We add those two results together, recursively calculating the crosssum of the truncated number.

Testing with Your Example #

Calling crosssum 12 will:

• First check 12 > 9, so compute 2 + crosssum 1
• Then crosssum 1 hits the base case, returns 1
• Final result: 2 + 1 = 3, which is correct.

Why Your Original Code Crashed #

If you got a thread killed error, it's likely you missed the base case for single-digit numbers, or had a wrong condition (like using n > 0 instead of n <=9 without handling the single-digit stop). Without that base case, the function keeps calling itself on smaller and smaller numbers until the stack overflows.

内容的提问来源于stack exchange，提问作者gewissen