Python non-default argument follows default argument

Question

Problem:

I am trying to implement the Pythagorean formula in my program using a function. But the program is not running for some reason. I am attaching my code snippet and the error message below:

from math import sqrt
def pythagorean (a, b=-1, c):
    return sqrt(a ** 2 + c ** 2)
pythagorean(a=3,c=9)

Error message:

  File "main.py", line 17
    def pythagorean (a, b=-1, c):
SyntaxError: non-default argument follows default argument

As you can see in the error message, non-default argument follows default argument. What is this error? How can I fix the error? Thanks for your kind concern buddy.

Answer 1

As I can see, you have a non-default argument after the default argument inside your function

def pythagorean (a, b=-1, c)

 the right way to write the function argument is:

def pythagorean (a, c, b=-1)

Moreover, you were wrong in your code on the line return function line. This is not gonna return anything. To fix the program, set all the arguments to default in your function. Now, the question could be, how do we know which two parameters are passed into the function? To make it sure we may have if-elf statement inside our function and end of the function we’ll get the result.

from math import sqrt
def pythagorean (a = -1, b = -1, c = -1):
    if a == -1:
        return sqrt(c**2 - b**2)
    elif b == -1:
        return sqrt(c**2 - a**2)
    elif c == -1:
        return sqrt(a**2 + b**2)
    else:
        return 0
pytho = pythagorean(a=3, c=9)
print pytho

This may help you. Good Day!