#include <stdio.h> int main() { int n; printf("Input an integer\n"); scanf("%d", &n); if (n & 1 == 0) printf("even\n"); else printf("odd\n"); return 0; } This program doesn't enter the If loop and always prints 'odd'. I understand that if(False) or if(0) is breaking condition but "n&1==0" is a TRUE condition for even numbers right? or am I missing something here?.
Enable all warnings in your compiler. Mine says:
warning: & has lower precedence than ==; == will be evaluated first [-Wparentheses] note: place parentheses around the & expression to evaluate it first The equality operators == and != have a higher priority than the bitwise AND operator.
So the condition in the if statements is equivalent to the following
if (n & ( 1 == 0 ) ) as 1 is not equal to 0 then the condition can be further rewritten like
if (n & 0) Thus the substatement of the if statement is never executed because n & 0 always evaluates to false ( 0 ).
To escape the logical mistake you could exchange the if and else statements.
if (n & 1 ) printf("odd\n"); else printf("even\n"); Take into account that according to the C Standard the function main without parameters shall be declared like
int main( void ) Here is a demonstrative program.
#include <stdio.h> int main(void) { while ( 1 ) { int n; printf( "Input an integer (0 - exit): " ); if ( scanf( "%d", &n ) != 1 || n == 0 ) break; printf( "%d is %s\n\n", n, n & 1 ? "odd" : "even" ); } return 0; } Its output might look like
Input an integer (0 - exit): 10 10 is even Input an integer (0 - exit): 9 9 is odd Input an integer (0 - exit): 8 8 is even Input an integer (0 - exit): 7 7 is odd Input an integer (0 - exit): 6 6 is even Input an integer (0 - exit): 5 5 is odd Input an integer (0 - exit): 4 4 is even Input an integer (0 - exit): 3 3 is odd Input an integer (0 - exit): 2 2 is even Input an integer (0 - exit): 1 1 is odd Input an integer (0 - exit): 0 
scanf( "%d", &n ) != 1 in your code as break condition?& has a surprisingly low precedence so you need (n & 1) == 0. Better still, use
if (n & 1) printf("odd\n"); else printf("even\n"); Note that this is implementation defined for negative int values due to differing complementing schemes, so you may well not get a correct answer for a negative n. Perhaps use unsigned?
you do not have to compare to zero. Any non zero value is the true, zero is false.
So it is enough to:
if (n & 1 ) printf("odd\n"); else printf("even\n"); or
printf("%s\n", (n&1) ? "odd" : "even"); 
(n & 1 == 0)->((n & 1) == 0)n & 1is zero,nis even.