An Array of One

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10 comments, last by CyberSlag5k 20 years, 1 month ago
Bah, you''re ruining my fun
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Awsome, everything is working correctly now. The answer was indeed to do a "tempBlock = new Block()" for each array element. I''m not sure what I was doing before, but I could get "blockArray[arrayCount].setPos(i*80, j*25)" to compile now.

Thank you to all who helped!

Mike

When you find yourself in the company of a halfling and an ill-tempered Dragon, remember, you do not have to outrun the Dragon...
Without order nothing can exist - without chaos nothing can evolve.

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