Can anyone explain this code?

Question

def func1():
Pass
def func2(n):
if n%2==0:
return none
print(func1() == func2(6))
print(func1() == func2(3))

https://code.sololearn.com/cLrH4cJd6NS0/?ref=app

Answer: true , true

But how???????

Answer 1

'func2' is defining like that :
def func2(n) :
if n%2 == 0:
return None
else:
return None

And your first function is the same as :
def func1() :
return None

print(func1()) # None
print(func2(2)) # None
print(func2(1)) # None

As you can see, your two functions are returning None, whatever happens.
>>> None == None
True
So, 'func1() == func2(a_value)' will be True in any case.

Answer 2

Théophile Thanks! Now I understood 🤗

Answer 3

An empty return statement is equal to :
'return None'

If a function does not return any value, it returns None.

import dis
def func():
pass
did.dis(func)

You can see :
LOAD_CONST 0 (None)
RETURN_VALUE

Answer 4

Théophile

func1() is already return none

but in the last line,
3 is argument of func2()

And in the func2() , there is a if statement

if 3%2==0 it's return none

And if this statement is true
Then,
Last line condition will become true and it prints true

But in the line no. 4 ,
3%2==1

So it condition will become false

But answer is true

but how?????

Answer 5

Théophile
But in line no. 4,

3%2==1

So condition will become false

Answer 6

The call 'func2(3)' will return None. The reason is simple : if n is not even, then the function returns nothing... So it returns None.

def func(a):
if a == 3:
return None
print(func(3)) # None
print(func(1)) # None

[EDIT]
'func1' will always return None, like 'func2'. That's why the result is always True.