python if/else & return

[lewashby]

Below is a short program I'M working of from udacity.com but I can't figure out why I'M getting the value 6 from the first call to the function.

# Define a procedure, biggest, that takes three
# numbers as inputs and returns the largest of
# those three numbers.

def biggest(first, second, third):
    if (first > second and third):
        return first
    if (second > first and third):
        return second
    else:
        return third




print biggest(3, 6, 9)
#>>> 9

print biggest(6, 9, 3)
#>>> 9

print biggest(9, 3, 6)
#>>> 9

print biggest(3, 3, 9)
#>>> 9

print biggest(9, 3, 9)
#>>> 9

[woooee]

Set third to 0 (False) and see how that affects things. An "and" is 2 nested if statements:

if first > second:
    if third:

See 5.1. Truth Value Testing

[Ene Uran]

# Define a procedure, biggest, that takes three
# numbers as inputs and returns the largest of
# those three numbers.

def biggest(first, second, third):
    if first > second and first > third:
        return first
    if second > first and second > third:
        return second
    else:
        return third


# result is 9 in each case
print biggest(3, 6, 9)
print biggest(6, 9, 3)
print biggest(9, 3, 6)
print biggest(3, 3, 9)
print biggest(9, 3, 9)

You could do:

def biggest(first, second, third):
    return max(first, second, third)

[woooee]

To be complete, a solution that iterates over the tuple instead of using if statements, and can accomodate any length of numbers input.

def biggest(*nums_in):
    ret_num = nums_in[0]
    for num in nums_in:
        if num > ret_num:
            ret_num = num
    return ret_num

print biggest(3, 6, 7)
print biggest(6, 5, 3, 8, 2)
print biggest(9, 3, 6)
print biggest(3, 3, 10)
print biggest(11, 3, 11)