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    <p>You're including 1 and the <code>candidate</code> in your <code>candidate%divisor</code> tests, both of which will always return 0, so every number you test will appear prime.</p> <p>Instead of this:</p> <pre><code>for(divisor=1;divisor&lt;=candidate;divisor++) </code></pre> <p>do this:</p> <pre><code>for(divisor=2;divisor&lt;candidate;divisor++) </code></pre> <h3>update</h3> <p>There are too many things wrong with your code to list, but here are a couple anyways:</p> <ul> <li>you're checking the result of <code>candidate%divisor</code> outside the for loop</li> <li>by the time you check <code>k</code>, <code>k</code> is always <code>candidate%candidate</code>, or 0</li> <li>you only ever check <code>k</code> once, after the divisor loop</li> <li>don't <code>continue</code> at the end of your loop, that's what happens by default</li> </ul> <p>Look, here is some pseudo-code for the simplest possible implementation. Try to find where your code deviates from this:</p> <pre><code>for candidate = down to up // assume the candidate is primt prime = true // check all divisors from 2 to the candidate - 1, inclusive for divisor = 2 to candidate - 1 if candidate % divisor == 0 // divisor is a factor of candidate // candidate isn't prime, so we can stop checking prime = false break end if next divisor // if prime is still true, we successfully tested every number from 2..candidate // and found no factors if prime print "candidate {candidate} is prime!" end if next candidate </code></pre>
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    1. POAbout Prime Generation in "C" - What is wrong with my code ? -
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    1. COThanks for the psuedocode , your first suggestion doesn't make difference in execution . :( If you extend your proposal for the "for" loops i hope i can understand easily. Sorry for acting as dummy :)
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    2. COQuestion is solved, thanks
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