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Referencing arrays with a variable
I'm referencing arrays with a variable, like this...
arr[Var];
However, this isn't working. Can you not dynamically reference arrays? This is what I am doing (probably excessive I know, but just need to sort out what I can and can't do with arrays.)
var blocks = new Array ();
var k : int;
var k2 : int;
for(k=0;k<blocks.length;k++){
k2 = k++;
blocks[k] = f1;
print (blocks[k]);
blocks [k2] = f2;
dif=f1-f2;
print (dif);
Thanks for any advice!
What you're doing should work, but I'm not sure what you're actually trying to do! Can you explain what this code is actually supposed to do? Right now, by the looks of things, blocks will always be empty since you never add any members to it, and even if you did, it will set every even-indexed block to f1, and every odd-indexed block to f2, and then print the difference between f1 and f2 every time. Importantly, dif will always be the same value here. You haven't shown us where f1 and f2 are declared, either.
Also, are you sure you want to be using the Array() class for this, not just a builtin array?
Why will dif always be the same value? (assu$$anonymous$$g varying other values throughout the array.)
Answer by TheIBPoliceman · Dec 01, 2011 at 09:32 PM
So you've created an array called blocks, but it has no elements in it as of yet. So your code in the loop isn't even going to run as blocks.length will always return zero.
You're using JS arrays there, so you'd add new code with array.Add(), or perhaps Push() (check out http://unity3d.com/support/documentation/ScriptReference/Array.html) for methods at the end of the page. You also can't just index at a new value outside the array to add an element, use Add or resize the array with array.length, which would then allow you to index at that point.
Yeah, this is totally wrong, but I should have added more to my question. I've added a bunch of other code, but it's roughly 200 LOC and it would be hard to post all relevent sections. I just wanted to know that I was getting dif and everything correctly.
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