how to check given argument is int or not ?

This is what you want to do. Here, try will multiply the argument if it can be converted to an integer. In the case that it cannot be converted to an integer you will get a ValueError, we want to tell python what to do if we get the ValueError. So we use except ValueError: In our case we want to multiply the string by the length of the string.

def squared(argument):

    try:
        answer = int(argument) * int(argument)
        return answer
    except ValueError:
        answer = argument * len(argument)
        return answer

Sahil,

Googling for "python is argument integer" will give you several good answers to your question. I say this because you're going to find that you do that often, and it often leads to Stack Overflow. And often, you'll find well thought out answers with specific examples.

Sahil;

Welcome to Treehouse!

Based on your code, you are on the right track with just a few syntax issues. Let's have a look...

Task 1

This challenge is similar to an earlier one. Remember, though, I want you to practice! You'll probably want to use try and except on this one. You might have to not use the else block, though.

Write a function named squared that takes a single argument.

If the argument can be converted into an integer, convert it and return the square of the number (num ** 2 or num * num).

If the argument cannot be turned into an integer (maybe it's a string of non-numbers?), return the argument multiplied by its length. Look in the file for examples.

def squared(argument):

  argument = int(input("enter any argument "))

  while argument == int:
    try:
        answer = argument * argument
           print("answer is {} ".format(answer))
    except valueError :
        answer = argument * len(argument)
           print("answer is {} ".format(answer))

So your function is taking an argument, that's great. We don't need to reassign it's value though on the second line of code. while argument == int doesn't do anything since we don't have a variable named int, right? Further for this challenge we don't even need a while statement.

In your try... except statements, the challenge wants you to return an answer, not just print it out. You will also likely want to utilize the int() method that we used previously in the course to try to convert your argument into an integer.

Hopefully that get's you pointed in the right direction. Post back, however, if you are still stuck.

Happy coding,
Ken

I'm thinking this is what you are trying to accomplish, If you are using input to get your number to square you won't need to use a parameter. If you run this script it will ask for a number to square. If a number isn't entered it will print that a number wasn't entered and will return to the top of the function. Once a number is entered it will print the answer.

def squared():

    try:
        argument = int(input("Enter a number to square "))
    except NameError:
        print("You didn't enter a number!")
        return squared()

    answer = argument * argument

    print("Your answer is " + str(answer))

squared()

def squared(arg):
    try:
        if int(arg):
            return arg * arg
    except ValueError:
        return arg * len(arg)

Hi Ken,

I got a TypeError: can't multiply sequence by non-int of type 'str'. Am I missing something here? Since the if statement becomes false when I put a string it, shouldn't the except function execute?

This also work

def squared(num):
    try:
        return int(num) **2
    except ValueError:
        return num * len(arg)