can anyone explain me this arthimatic operations

Question

https://code.sololearn.com/cv8wdCgshySf/?ref=app

Answer 1

Granger
Great explanation
I didn't know about the byte shift

Answer 2

Mr. 12

When using the | operator, you are comparing the 2 sets against each other. Overlapping them in a way.
So when you overlap 0b1100 and 0b110, the following occurs as you look at the numbers from the left.
1 => 1
1 1 => 1
0 1 => 1
0 0 => 0

Result is 0b1110

Remember that 1 is true & 0 is false, so if given an option to choose between true and false, the code will return true

print(2|3) #3
print(0b10|0b11) #3 =>0b11

Answer 3

Mr. 12
Go through this section to see how set operators work

https://www.sololearn.com/learn/JUMP_LINK__&&__Python__&&__JUMP_LINK/2464

Answer 4

I have seen something like this before which was explained as set operators working on the binary format of the numbers.

The following snippet may help you understand more

print(3|5)
print(bin(3))
print(bin(4))
print(bin(5))

Answer 5

Mr. 12
Ahh, the joys of binary numbers.
As you discovered, 0b10 is the binary equivilent of 2.

Therefore:
0b11 is 3
0b100 is 4
0b101 is 5

print(0b11|0b101) is 0b111
0b111 is 7

Answer 6

Rik Wittkopp what if one contain 01 and another contain 0011?

Answer 7

Rik Wittkopp bro, i didn't understand

print (bin(2)) =01b0
how and why?

Answer 8

Rik Wittkopp but how bro
0b11 is 3, and 0b101 is 5
but how it becomes 7?

and also what will &, |, <<, >>, ^ will do?

Answer 9

Granger thank you bro, i know about binary number but i confused after it is printing 0b11 for 3,which is only 11 or 011 or 0011 for 3 in binary.
i just confused with 'b' presenting in 0b11.

And also thank your for explaining the arthimatic operators, Now I Got It.

Answer 10

Granger so, Shifting means just adding zeors to left of right?

Answer 11

Rik Wittkopp can u explain me more about this?

if we join 2 | 3 (0b10 | 0b11)
ans would be 0b10 right?
because 1 is already there in 0b10?

Mr. 12
Ahh, the joys of binary numbers.
As you discovered, 0b10 is the binary equivilent of 2.

Therefore:
0b11 is 3
0b100 is 4
0b101 is 5

print(0b11|0b101) is 0b111
0b111 is 7

Answer 12

Mr. 12
0 => 0
0 => 0
1 0 => 1
1 1 => 1

Result => 0b11

Same rules apply