I want explanation for this code, please.

Question

Int n;
Scanf("%d", & n) ;
If(n&0x01)
{ Printf("odd") ; }
Else
{ printf("even") ; }

Answer 1

0x01 --> means 01 in hex . So, the number is : 00000001
& means bitwise operator

As a result, if n=10 for example in dec we have

00000 1 10
& 00000001
--------------
--> returns 0

if n=11 for example in dec we have

00000 1 11
& 00000001
--------------
--> returns 1

So, depending on the last digit of n the output will be 1 or 0 and the decision will be odd or even.

-- ==================
The statement is used for performance reasons as it is much faster than n%2

Answer 2

that doesn't seems to be valid logic

& is AND bitwise operator. so for example if n is 2 the operation is

1 0 => 2
0 1 => 1
_____&
0 0 => 0 (0 in c/c++ is false)
in this case the logic is valid

but, lets say the input is 3
1 1 => 3
0 1 => 1
_______&
0 1 => 1 (true, since non zero is true)

the result are true, but 3 is an odd number

Answer 3

This program :
- takes an input from user (integer)
- test with binary operator and (&) if this number is odd or even : all odd numbers, in binary, finish by bit 1 (like 3 : 0011) and all even number finish by 0 bit (like 2 : 0010). So if number is odd, n&0x01 is 1, else, it is 0.