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Krocker

Sun Oct 20, 2013 5:13 pm


odd indices of an array

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hi, so i need to output the odd number indices of an array without using any if statments. i know that odd = 2n+1 where n is the an integer starting at 0. my trouble is getting the for loop im using to stop when all the odd indices are  outputed. So far, all i get it to is before the last index or beyond the last index, giving me an out of bound error. This is my for loop so far:

[code]
System.out.println ("\n\nValues at odd-numbered indices:");
		int odd = 2;
		int shiftedOddArray = 0;
		
		for (int i = 0; i < (odd-1); i++){
			odd = (2*i)+1;
			shiftedOddArray = shiftedArray[odd];
			System.out.print ( shiftedOddArray + "\t"); 
		}
[/code]

shiftedArray is the original array that i have to output every odd index. The number of indices and the numbers in the indices are all determined by the user at the begining of the program.


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Raknarg

Sun Oct 20, 2013 7:47 pm


RE:odd indices of an array

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You don't have to use i++. THat's just a shortcut for saying add one to i. You can do whatever you want in there (I'm pretty sure)


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Tony

Sun Oct 20, 2013 8:48 pm


RE:odd indices of an array

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anything you want, or even nothing.
[code]
for(;;){
   ...
}
[/code]
is a valid loop. At least for C, probably Java.


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Krocker

Sun Oct 20, 2013 9:42 pm


RE:odd indices of an array

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but how would that stop when the program has gone through the number of odd indices that are possiible according tho the user specified number of indices?


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Insectoid

Sun Oct 20, 2013 10:13 pm


RE:odd indices of an array

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How would you do it if you wanted to run through all of the indices of the array up to the user specified limit?


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Raknarg

Mon Oct 21, 2013 8:05 am


RE:odd indices of an array

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@Tony is that a loop that does nothing or loops infinitely?


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Insectoid

Mon Oct 21, 2013 8:14 am


RE:odd indices of an array

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The 2nd argument to a for loop is the exit condition. If there is no exit condition, what do you think will happen?


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Krocker

Mon Oct 21, 2013 8:36 am


RE:odd indices of an array

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if there isnt an exit condition, there will be an infinite loop, thats not what im looking for. Anyways, i figured it out, i simply used a while loop and said that while (odd< numberOfArray)


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C14

Tue Oct 22, 2013 1:04 pm


Re: RE:odd indices of an array

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Anyways, i figured it out, i simply used a while loop and said that while (odd< numberOfArray)

Can we see your code?


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Krocker

Fri Oct 25, 2013 8:28 am


Re: odd indices of an array

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Heres the code that i used:

[code]
                System.out.println ("\n\nValues at odd-numbered indices:");
		int odd = 1; //declare and initializes variables to be used 
		int shiftedOddArray = 0;
		int n = 0;

		//does through each odd indices and outputs the value in order of the shifted array by calling shiftedArray [odd] where odd is 2n+1
		while (odd