英文:
Recursive sumNumbers in Haskell
问题
findSum :: Int -> Int -> Int -> Int
findSum a b n
| a >= 0 && b > 0 && n > 3 = findSum1 a b n + findSum2 a b n + findSum3 a b n
| otherwise = 0
where
findSum1 :: Int -> Int -> Int -> Int
findSum1 a b n
| n <= 3 = 0
| otherwise = a + 2 ^ (n-1) * b + findSum1 a b (n-1)
findSum2 :: Int -> Int -> Int -> Int
findSum2 a b n
| n <= 3 = 0
| otherwise = a + 2 ^ (n-2) * b + findSum2 a b (n-2)
findSum3 :: Int -> Int -> Int -> Int
findSum3 a b n
| n < 0 = 0
| otherwise = a + 2 ^ (n-3) * b + findSum3 a b (n-3)
英文:
My task is: to write a function who calculate (a + 2^0 * b + 2^1 * b + 2^2 * b until it happens 2^(n-1)*b), other who calculate (a + 2^0 * b + 2^1 * b + 2^2 * b until it happens 2^(n-2)*b) and another one who calculate (a + 2^0 * b + 2^1 * b + 2^2 * b until it happens 2^(n-3)*b).
I have to write a function who calculate sum of the three functions but this function has to accept 3 argument - the numbers a b n!
n is always greater than 3!
This is what I have done until this moment:
findSum :: Int -> Int -> Int -> Int
findSum a b n
| a >= 0 && b > 0 && n > 3 = findSum1 a b n + findSum2 a b n + findSum3 a b n
| otherwise = 0
where
findSum1 :: Int -> Int -> Int -> Int
findSum1 a b n
| n <= 3 = 0
| otherwise = a + 2 ^ (n-1) * b + findSum1 a b (n-1)
findSum2 :: Int -> Int -> Int -> Int
findSum2 a b n
| n <= 3 = 0
| otherwise = a + 2 ^ (n-2) * b + findSum2 a b (n-2)
findSum3 :: Int -> Int -> Int -> Int
findSum3 a b n
| n < 0 = 0
| otherwise = a + 2 ^ (n-3) * b + findSum3 a b (n-3)
But It gives me an error:
Exception: Negative exponent.
答案1
得分: 2
findSum3 在第一个条件中比较的是 n < 0 而不是 n <= 3。
这导致它试图计算 2 ^ (n-3)，而实际上 n < 3，最终导致负指数，这是不允许的。
英文:
findSum3 does compare n < 0 instead of n <= 3 in the first clause.
That leads to it trying to calculate 2 ^ (n-3) with an n < 3 eventually which results in a negative exponent, and that's not allowed.
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