0
I want explanation for this code, please.
Int n; Scanf("%d", & n) ; If(n&0x01) { Printf("odd") ; } Else { printf("even") ; }
3 Réponses
+ 2
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
+ 5
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
+ 1
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.