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 vignesh1988i 23Sep2009 08:55

C programming

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 )??????????:):):):):):)

Re: C programming

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

Code: CPP

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

 vignesh1988i 25Sep2009 02:51

Re: C programming

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

Re: C programming

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

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

 vignesh1988i 25Sep2009 18:17

Re: C programming

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

thank u

Re: C programming

Then I would do it this way ;)

Code: CPP

`char Result[2][5] = {"EVEN", "ODD"};printf("%d is %s\n", A, Result[A&1]);`

 vignesh1988i 26Sep2009 09:46

Re: C programming

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

 xpi0t0s 26Sep2009 12:23

Re: C programming

Code:

`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 26Sep2009 12:30

Re: C programming

hmmmmmm this is the right answer..... :)

 Originally Posted by xpi0t0s (Post 58004) Code: `printf("1 for odd, 0 for even -> %d \n",A&1);`