Why the output is different from my answer? [Solved]

Question

arr = [1,2]
def func(n,a):
if(a is not arr):
a.append(n)
return list(a)
a = arr
a = func(3,a)
print(a)

Output - [1,2]
Why not - [1,2,3]

Answer 1

SauravK.__
Because outside the function the 'a' has changed after the first call.

a = arr # <--- a is arr

# When we called the function in this line, the 'a' still holds the reference of 'arr', but after we called the function, the 'a' now holds a different value.
a = func(3, a)

a = func(4, a) # <--- a is not arr

a = func(5, a) # <--- a is not arr

Notice that the value of 'a' is changing everytime we call the function and assigning the returned value to 'a',.

Answer 2

The condition "if (a is not arr)" inside the function is False and therefore it will not append 3 to `a`.

`a` and `arr` has the same memory address as we explicitly declared `arr` to `a` in line 6, i.e. they are the same object.

Remember that when assigning list to another variable, we don't pass its values, we pass its reference.

Answer 3

I'm not sure if I understand the 'check condition one by one'.

Though I can say that yes it will check the condition when we call the function.

First, we call the function on (2) and the variable `a` now holds the "new" returned list.
And so on (2) the value of `a` has been changed and so the next time we call the function in (3), the condition will now become True.

Remember that [1, 2] is not [1, 2] if they have different memory address.

For example:
x = [1, 2]
y = [1, 2]

print(x == y) # True
print(x is y) # False

Answer 4

Yes it will, I did not notice that the variable that uses the function has the same variable 'a'.

a = arr
a = func(3, a) # <--- a is arr

a = func(4, a) # <--- a is not arr because a is now a different [1, 2] returned by the function therefore the condition will become True and append 4.

Things will be different if we use variables except 'a'.

a = arr
x = func(3, a)
y = func(4, a)
z = func(5, a)

>> [1, 2]

Because the value of 'a' did not change and it still stores the reference of 'arr'.

Answer 5

SauravK.__

Before the first function call 'a' was arr so if you check (a is not arr) it will return False and not append value but now returned value's reference is changed so after 1st function call if you again check (a is not arr) then it will return True and append 4 in 'a'

If you assign arr to 'a' after 1st function call then 4 will not be append in 'a'

You can try this

arr = [1, 2]

def func(n, a):
if(a is not arr):
a.append(n)
return list(a)

a = arr
a = func(3, a)

print (a)
a = arr #assigned again.

#try to remove the above line and see the effect

a = func(4, a)
print(a)

Answer 6

SauravK.__

Question is correct but problem is in if condition.

if (a is not arr) here this condition will be false because a is a arr because you have assigned arr to a and passed a in function so reference of a and arr will be same.

Answer 7

Cyan Finally ! doubt cleared
Thank you both of u
in short :

a = arr --> a is arr
a = func(3,a) = arr --> a is arr
a = func(4,a)
where a = func(3,a)=arr
a!=func(4,a) also a is not func(4,a) --> a is not arr
now if statement is correct & it will append n
after a

Answer 8

Cyan what will happen in case if we add " a = fun(4,a) " after 7th line?

Answer 9

Cyan I also felt the same but when I tried this in code editor then output comes : [1,2,4]

Answer 10

SauravK.__

There should be

if (a is arr):
a.append(n)

Answer 11

Cyan you want to say that:
first it will check condition one by one?
1.a = arr
2.a = func(3,a)
3.a = func(4,a)

Answer 12

I Am AJ ! but I feel question is correct. dont you?

Answer 13

I didn't get your point clearly but I wrote whatever given in that question..

basically this was the question:
arr = [1,2]
def func(n,a):
if(a is not arr):
a.append(n)
return list(a)
a = arr
a = func(3,a)
a = func(4,a)
print(a)

I was stuck here
I didn't understand that why the condition of if statement is true for a = func(4,a) but not for a = func(3,a)

Answer 14

I Am AJ ! Bhaiya this is same what I tried earlier !
output will be [1,2]
as a = arr also a = func(3,a)
so arr =func(3,a) --> a is arr..