Why this code isn't work?

Question

def a3():
x=int(input('Enter Number:-'))
if (x>50):
print ("yes")
else:
print ("no")

https://code.sololearn.com/cggxsW3M9760/?ref=app

Answer 1

Subod Hiruna try this
input=int(input('Enter Number:-'))
def a3(x):
print(x)
if (x>50):
print ("yes")
else:
print ("no")
a3(input)

Answer 2

Subod Hiruna ,

there is no output because you have defined a function a3() , but you never have called it.

put a3() as the last line of your code without indentation. this will call your function.

Answer 3

You need to call function

Answer 4

Atul [Inactive] ,

to call the function we need to write a3(), otherwise the function will not be executet.

Answer 5

Lothar yes that's what I told. And I think it could be understood by Subod Hiruna

Answer 6

This works easily 🤷🏻‍♂️

x=int(input("Enter Number"))
y=int(50)
if x>y :
print("yes")
else:
print("no")

Answer 7

https://code.sololearn.com/cBEE7fCXIp2g/?ref=app

Use this

Answer 8

At the bottom you just write a3

Answer 9

Subod Hiruna Your code prints nothing as you've not added the extra piece of code to execute the function 'a3'. 'a3' is a variable in Python that stores the reference to the function object in the heap. A pair of parentheses after this reference calls the function. Here's how you can make it work:

def a3():
# Uhm, you might as well could deal without the extra variable 'x'
x = int(input("Enter the number: "))
if x > 50:
print("yes")
else:
print("no")

Here's an even shorter version of your code:

def a3(): print("yes" if int(input("Enter: ")) > 50 else "no")

# I hope that this answer helps you with your query. Happy coding!

Answer 10

Because you did not call the function
It should be like a
a3()

Answer 11

Call the function