Print(2*5-2%1) what to do first? Tell me whole process.

Answer 1

first we have will see which operators have high precedence
In the question, (*,%) has high precedence than (-)
(*,%) has equal to precedence so we will use it associativity rule i.e. left to right so the operator from (*,%) will be solved first that is one the left side

We will solve it like this-
(2*5-2%1)
(10-2%1) * is on the left side so it got first solved before %
(10-0) % has more precedence than -
10 10 is the final answer

Hope you got understand........

Answer 2

It is actually following the rule of BODMAS...

Brackets
Orders
Division
Multiplication
Addition
Subtraction

Here Print(2*5-2%1) has no bracket so passes to division method i.e. 2%1 = 0 then multiplication 2*5 = 10 then at last substraction of 10-0 which is 10 ....so it will print 10...

Answer 3

Ben sidhom Akram Bro firstly division is done then multiplication.....

Answer 4

1. 2*5=10
2. 2%1=0
3. 10-0=10

Answer 5

* and % have the same precedence and their precedence is higher than that of -
Therefore this is evaluated as follows, (2*5)-(2%1) which gives 10.

Answer 6

Thanx to all.

Answer 7

Firstly , do the multiplication of 2*5=>10
Secondly , do the modulo of 2%1=>0
Finally do the subtraction =>10-0=10