函数的返回值为非

Question

我是Python的初学者。下面是我用python编写的一个简单代码，它递归地计算字符串中某个字符的出现次数。输入是硬编码的。如有需要，请参阅评论

# . : Logic : .
# Recursively counts the number of occurrences of a character
# in a given string and returns the count when the length of the string becomes zero

def recSearch(g_str, g_ch, val):
    # Length of string is zero, hence function will terminate
    if len(g_str) is 0:
        x = val
        print "Number of times [%s] occurs is %d" % (g_ch, x) 
        return val
        # ERROR : Returning none instead of a number

    # 1st character of string is a match
    # Hence val is incremented 
    elif g_str[0] is g_ch:
        recSearch(g_str[1:], g_ch, val + 1)

    # 1st character of string is NOT a match
    else:
        recSearch(g_str[1:], g_ch, val)

strSer = "this is most probably a valid string"
charSer = "t"
# Answer should be 3 for this input
# Feel free to experiment

print "The input string = [%s]" % (strSer)
print "Character to be found = [%s]" % (charSer)

i = recSearch(strSer, charSer,0)

print "I should be getting a valid return value %d" % i
# But instead, I'm getting an error here

但我得到以下错误

Traceback (most recent call last):
File "sample2.py", line 31, in <module>
print "I should be getting a valid return value %d" % i
TypeError: %d format: a number is required, not NoneType

原因是什么？你知道吗

Answer 1

代码在elif、else块中的递归调用中缺少return；返回不带return语句的函数会导致None返回。你知道吗

elif g_str[0] is g_ch:
    return recSearch(g_str[1:], g_ch, val + 1)  # < -
else:
    return recSearch(g_str[1:], g_ch, val)  # < -