返回 None 而不是 False

Question

I have this code that should return false at the end of the function if one of the condition isn't met, but the output keeps saying None, why so?

def check_largest_and_smallest():
    case1 = largest_and_smallest(17, 1, 6)
    case2 = largest_and_smallest(1, 16, 6)
    case3 = largest_and_smallest(1, 1, 2)
    case4 = largest_and_smallest(1, 1, 1)
    case5 = largest_and_smallest(-3, -4, 0)
    if case1 == (17, 1):
        if case2 == (17, 1):
            if case3 == (2, 1):
                if case4 == (1, 1):
                    if case5 == (0, -4):
                        return True
    else:
        return False

The largest_and_smallest function is:

def largest_and_smallest(num1, num2, num3):


    largest = None
    smallest = None

    if (num1 >= num2) and (num1 >= num3):
        if num2 <= num3:
            largest = num1
            smallest = num2
        else:
            largest = num1
            smallest = num3
    elif (num2 >= num1) and (num2 >= num3):
        if num1 <= num3:
            largest = num2
            smallest = num1
        else:
            largest = num2
            smallest = num3
    elif (num3 >= num1) and (num3 >= num2):
        if num1 <= num2:
            largest = num3
            smallest = num1
        else:
            largest = num3
            smallest = num2

    return largest, smallest

Answer 1

The reason is due to the use of else : as soon as the first if statement is checked, there's no else for the other situations, so the function simply returns None .

    case1 == (17, 1)
    case2 == (16, 1)

    if case1 == (17, 1):
        if case2 == (17, 1):
        # here case2 doesn't match, but there's no corresponding else!
    else:
        return False

A simple fix is to just return False without any else :

    if case1 == (17, 1):
        if case2 == (17, 1):
            if case3 == (2, 1):
                if case4 == (1, 1):
                    if case5 == (0, -4):
                        return True
    return False

Or, even better, use all( ) which checks all iterable items and returns False as soon as one of them is not True , meaning that the function will stop evaluating the following items whenever that happens:

    return all((case1 == (17, 1), case2 == (17, 1), case3 == (2, 1),
        case4 == (1, 1), case5 == (0, -4)))

Answer 2

At a glance, for me, the indentation for check_largest_and_smallest() is not correct. Then, you should not use else, just return in the end will satisfy your intended logic.

def check_largest_and_smallest():
    case1 = largest_and_smallest(17, 1, 6)
    case2 = largest_and_smallest(1, 17, 6)
    case3 = largest_and_smallest(1, 1, 2)
    case4 = largest_and_smallest(1, 1, 1)
    case5 = largest_and_smallest(-3, -4, 0)
    if case1 == (17, 1):
        if case2 == (17, 1):
            if case3 == (2, 1):
                if case4 == (1, 1):
                    if case5 == (0, -4):
                        return True

    return False

def largest_and_smallest(num1, num2, num3):
    """
    Takes 3 numbers as arguments and returns
    the largest number and smallest number among them.
    """
    largest = None
    smallest = None

    if (num1 >= num2) and (num1 >= num3):
        if num2 <= num3:
            largest = num1
            smallest = num2
        else:
            largest = num1
            smallest = num3
    elif (num2 >= num1) and (num2 >= num3):
        if num1 <= num3:
            largest = num2
            smallest = num1
        else:
            largest = num2
            smallest = num3
    elif (num3 >= num1) and (num3 >= num2):
        if num1 <= num2:
            largest = num3
            smallest = num1
        else:
            largest = num3
            smallest = num2

    return largest, smallest


check_largest_and_smallest()

Answer 3

Consider what happens if case1 is true but case2 isn't.

Assuming that if ladder indentation is what you intend, you're only returning false if case1 isn't true, but if case1 is true and any of the other cases are false, your code returns nothing.

Answer 4

Try this:

if case1 == (17, 1) and case2 == (17, 1) \
      and case3 == (2, 1) \
      and case4 == (1, 1) \
      and case5 == (0, -4):
    return True
else:
    return False

Answer 5

For the minimum/maximum function, try something along the lines of:

def minmax(*x):
    return min(x),max(x)

In any case, you can always write

a < b < c

instead of

a<b and b<c

Answer 6

 if check_largest_and_smallest == False:
        print ("false") 
 else:
        print ("true")

with that you can also check the value of the Boolean:) I liked that I could learn also how Python works with interrogates like that. So thanks to the ppl. that solved it:)