Can you please explain the output of this code?

Question

def chek(x):
if x%3==0:
return False;
return True;
a = sum([i for in range(6) if chek(i)]);
print(a)

Answer 1

The chek() function checks if the given number is not divisible by 3 (returns False if x is divisible by 3 and True otherwise).

[i for i in range(6) if chek(i)] creates list from numbers 0, 1, 2, 3, 4, 5 (range(6)) filtering them through function chek() (numbers that are not divisible by 3).
The list will be [1, 2, 4, 5]

sum returns the sum of that numbers 1+2+4+5 = 12

print outputs the result.

Bur your code will not compile because you forgot to put i between "for" and "in"
line
a = sum([i for in range(6) if chek(i)]);
should be:
a = sum([i for i in range(6) if chek(i)]);

Answer 2

The explanation is in the code

https://code.sololearn.com/cRr8WXlW2C5G/?ref=app

Answer 3

WhyFry Yes but can you explain?

Answer 4

The Sylar yes but what that return True and False doing there with different indentation?

Answer 5

def chek(x):
if x%3==0:
return False;
return True;
b = sum([i for i in range(6) if chek(i)])
print(b)

About the above code in short

if the number in the range(6) have a zero remainder on dividing with 3 (Basically a multiple of 3) then it is not added in the list because the function chek(i) will return False for that.

So the list will be

b = [1, 2, 4, 5]

b = sum([1, 2, 4, 5]) gives 12

Basic maths... Just add all elements

Answer 6

Search 'list comprehension'.
Basically, from range 0 to 5, it check which number is not divisible by 3.
So they will be 1,2,4,5.
And sum of them is 12.
That is ! :)

Answer 7

chek(0) will get false, right?
So a will become:
a = [i for i in range(6) if false]

So, 0 will not be inserted because 'if' statement is false.
So do the 3.
But 1,2,4,5 will be included in the list because 'if' statement become true from chek(1 or 2 or 4 or 5).
Hope it's clear now.