爲什麼參數類型由不同的比較運算符改變？

Question

當我處理Ackermann function：

def ackermann(m,n): 
    if m == 0: 
    return n + 1 
    if m > 0 and n == 0: 
    return ackermann(m-1,1) 
    elif m > 0 and n > 0: 
     return ackermann(m-1,ackermann(m,n-1)) 

我寫錯運營商在第二if聲明：

if m == 0 and n == 0: 
    return ackermann(m-1,1) 

但我得到了一個錯誤：

File "C:/Users/jacob/Desktop/think python/exercise 6-2.py", line 15, in ackermann 
return ackermann(m-1,ackermann(m,n-1)) 

File "C:/Users/jacob/Desktop/think python/exercise 6-2.py", line 14, in ackermann 
elif int(m) > 0 and int(n) > 0: 

TypeError: unorderable types: NoneType() > int() 

爲什麼參數類型的改變運營商？

Answer 1

行，從而產生錯誤的代碼是通過一個例子

def ackermann(m,n): 
    if m == 0: 
    return n + 1 
    if m == 0 and n == 0: 
    return ackermann(m-1,1) 
    elif m > 0 and n > 0: 
     return ackermann(m-1,ackermann(m,n-1)) 

讓我們步驟。

ackermann(2, 1) 
    # m = 2, n = 1 
    m == 0: False 
    m == 0 and n == 0: False 
    m > 0 and n > 0: True 
    return ackermann(m-1,ackermann(m,n-1)) 
    ackermann(m,n-1) # the inner call 
    ackermann(2,1-1) 
     # m = 2, n = 0 
     m == 0: False 
     m == 0 and n == 0: False 
     m > 0 and n > 0: False 
     # None of the conditions matched, so we (implicitly) return None 
     # because we didn't execute an explicit return 
    ackermann(2-1,None) # the outer call 
     # m = 1, n = None 
     m == 0: False 
     m == 0 and n == 0: False 
     m > 0 and n > 0: 
     m > 0: True # m == 1 
     n > 0: error! # n == None 

Answer 2

我懷疑你沒有達到任何一個條件，並且會導致隱式返回值爲None的函數結束。

嘗試在函數結尾處返回一些有效值。

例如，

def ackermann(m,n): 
    if m == 0: 
    return n + 1 
    if m > 0 and n == 0: 
    return ackermann(m-1,1) 
    elif m > 0 and n > 0: 
     return ackermann(m-1,ackermann(m,n-1)) 

    # Writing an explicit return to make sure we always 
    # return some integer and not None. Change this accordingly 
    return 1