[vignesh1988i]

write a C program , to find whether a number is ODD or EVEN without using Control structures , looping, arrays, functions , operators(namely relational, arithmetic , logical )??????????

[SaswatPadhi]

OK, we can't use relational, arithmetic, logical operators, but we can use bit-wise and operator.

if(A&1) {printf("%d is odd\n", A);}
else      {printf("%d is even\n", A);}

[vignesh1988i]

but control structures(if... else) also should not be used...... but ur logic is right....
try that!!!!!

[SaswatPadhi]

We can always replace the if...else with the conditional operator <condition> ? <true case> : <false case> .
That way, the code becomes :

(A&1) ? printf("%d is odd\n", A) : printf("%d is even\n", A);

[vignesh1988i]

hmmmm then i ll put my question , in this way... there is a way to print without this conditional operator also.... try

thank u

[SaswatPadhi]

Then I would do it this way

char Result[2][5] = {"EVEN", "ODD"};

printf("%d is %s\n", A, Result[A&1]);

[vignesh1988i]

hmmmm ..in my question i already said without using arrays toooo u should do it!!!!!!!

[xpi0t0s]

printf("1 for odd, 0 for even -> %d \n",A&1);

If that still doesn't answer, please copy and paste the ORIGINAL question, not your summary of what you think it says.

[vignesh1988i]

hmmmmmm this is the right answer.....

[SaswatPadhi]

Quote:

Originally Posted by vignesh1988i

hmmmm ..in my question i already said without using arrays toooo u should do it!!!!!!!

My mistake

Quote:

Originally Posted by xpi0t0s

printf("1 for odd, 0 for even -> %d \n",A&1);

Nice.