Display only odd number in arrary 1..10

[nelsonkkyy]

What is it you are trying to achieve?
<Replace all the <> with your answers/code and remove the <>>


What is the problem you are having?
<Answer Here>


Describe what you have tried to solve this problem
<Answer Here>


Post any relevant code (You may choose to attach the file instead of posting the code if it is too long)
<Am i doing it right?>

Turing:

var num:array 1..10 of int for i:1..10 mod put num(i) end for %How to do it???!!!!!?&

Please specify what version of Turing you are using
<Answer Here>

[DtY]

So far you're good. You're currently displaying every number in the array, now you just need to filter it so it only shows the numbers that are odd.

You'll need
- a conditional statement (if)
- The modulo (mod) operator

[nelsonkkyy]

Can anyone show me how to write it?????/

[Zren]

Turing:

for i:1..9 by 2 put num(i) end for

Or the simplier method... just pump the counter by 2 starting it out on an odd number.

[nelsonkkyy]

var num: array 1..9 of int
for i:1..9 by 2
put num(i)
end for



It cannot run.

[Zren]

Did you intialize the values of the variables before trying to output them?

[nelsonkkyy]

HOW to intialize the values of the variables??

[darkn00b]

So you need to give the nums a value. You can do this a few ways, on is:

var num: array 1..9 of int
for i:1..9
get num(i)
end for
for i:1..9 by 2
put num(i)
end for

This will allow you the person who runs the program to type in a value for all of the nums, and then it outputs only the odd ones.

[DtY]

Zren @ Thu Oct 22, 2009 6:06 pm wrote:

Turing:

for i:1..9 by 2 put num(i) end for

Or the simplier method... just pump the counter by 2 starting it out on an odd number.

That shows odd indices, not the odd numbers

[btiffin]

On Odd Even tests.

number AND 1

If the result is 0, the number is even.
If the result is 1, (low bit set), the number is odd.

Cheers

[andrew.]

Simple way:

Turing:

for i : 1..9 by 2     put i end for

Harder way:

Turing:

for i : 1..9     if not i mod 2 = 0 then         put i     end if end for

[DtY]

andrew. @ Sat Oct 24, 2009 7:26 am wrote:

Simple way:

Turing:

for i : 1..9 by 2     put i end for

Harder way:

Turing:

for i : 1..9     if not i mod 2 = 0 then         put i     end if end for

Again, he's looking for odd numbers in the array, not the items with odd indices, so the first one is not right, the second one should be if num(i) mod 2 = 1

[andrew.]

Oh I didn't understand the question. I thought he wanted the odd numbers in a range, not the odd numbers out of a list of numbers.

[btiffin]

Crusty Old guy again

Try and not use if num(i) mod 2 = 1, when [b]num(i) and 1 = 1[/i] will be more efficient.

It's not a huge deal, but it is a deal.
Some REBOL timings;
Using modulo and integer arithmetic

code:

>> dt [for i 0 100000 1 [unless equal? 0 MOD i 2 [print i]]] ... 99993 99995 99997 99999 == 0:00:02.033888

delta time of 2 seconds.

Using logical and and bitwise 'arithmetic'

code:

>> dt [for i 0 100000 1 [unless equal? 0 i AND 1 [print i]]] 99993 99995 99997 99999 == 0:00:01.563786

delta time of 1.6 seconds.

So, if you write code that has to check 100,000 numbers for parity through your life time, using bit-wise you'll get like a whole extra half second of 'not waiting around' life to live.

Choose life, aim for efficient code.



Cheers

[DtY]

btiffin @ Sat Oct 24, 2009 11:31 am wrote:

aim for efficient code.

This might actually be relevant if it was posted anywhere but Turing help :p
But yeah, using bitwise and would be considerably faster because it doesn't actually need to divide, and learning optimizations like that will be helpful in the future