如何检查函数是否无返回值及Python无特质数字统计程序bug排查

Let's break down your two questions clearly:

In Python, if a function has no explicit return statement, or uses return without a value, it will default to returning None. To check if a function has "no return value" (i.e., returns None), you can directly compare its result to None:

def no_return_func():
    pass

result = no_return_func()
if result is None:
    print("This function has no return value")

Important distinction: Don't confuse returning an empty value (like '' or []) with having no return value (returning None). Your find_personality function always has a return value—when a number has no traits, it returns an empty string '', not None.

Your code's logic for counting dull numbers is flawed, let's fix that:

Root Cause #

Your check for dull numbers uses if '' in trts:—but in Python, an empty string '' is a substring of any string, whether trts is empty or has traits. For example:

print('' in 'odd')  # Outputs True
print('' in '')     # Also outputs True

This is why every number gets counted as dull.

Fixed Solution #

The correct check is to verify if trts is exactly an empty string with if trts == ''. I also fixed a minor grammar issue where empty traits would produce awkward output like "X is an number.".

Here's the revised full code:

arrogant_numbers = [3, 6, 7, 23, 25, 35, 39, 66, 68, 112, 119, 254, 259, 732, 737, 4565, 4663, 13330, 13730, 29880, 29998, 79670, 80015, 230054, 239068, 1534301, 1607352, 2060587, 21700891, 99167753, 99873125]
def find_personality(number):
    traits = ''
    # Find the traits of the number
    if number % 2 != 0:
        traits = traits + 'odd '
    # Check for other traits after this
    if number > 10000:
        traits = traits + 'excessive '
    if '3' in str(number):
        traits = traits + 'irksome '
    if number in arrogant_numbers:
        traits = traits + 'arrogant '
    return traits

# Write the rest of your program after this
dullcounter = 0
numbers = input('Enter numbers: ')
number_list = numbers.split()
for word in number_list:
    num = int(word)
    trts = find_personality(num)
    if trts:
        # Clean up trailing space for nicer output
        clean_traits = trts.strip()
        print(f"{word} is an {clean_traits} number.")
    else:
        print(f"{word} is a dull number.")
        dullcounter += 1
print(f'Dull numbers: {dullcounter}')

Extra Improvements #

1. Added strip() to remove trailing spaces from traits (so output looks like "X is an odd number." instead of "X is an odd number.")
2. Stored int(word) in a num variable for readability
3. Used if trts: instead of if trts != ''—empty strings evaluate to False in boolean checks, which is more Pythonic

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