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    <p>SUMMARY:</p> <pre><code>double roundit(double num, double N) { double d = log10(num); double power; if (num &gt; 0) { d = ceil(d); power = -(d-N); } else { d = floor(d); power = -(d-N); } return (int)(num * pow(10.0, power) + 0.5) * pow(10.0, -power); } </code></pre> <hr> <p>So you need to find the decimal place of the first non-zero digit, then save the next N-1 digits, then round the Nth digit based on the rest.</p> <p>We can use log to do the first.</p> <pre><code>log 1239451 = 6.09 log 12.1257 = 1.08 log 0.0681 = -1.16 </code></pre> <p>So for numbers > 0, take the ceil of the log. For numbers &lt; 0, take the floor of the log.</p> <p>Now we have the digit <code>d</code>: 7 in the first case, 2 in the 2nd, -2 in the 3rd.</p> <p>We have to round the <code>(d-N)</code>th digit. Something like:</p> <pre><code>double roundedrest = num * pow(10, -(d-N)); pow(1239451, -4) = 123.9451 pow(12.1257, 1) = 121.257 pow(0.0681, 4) = 681 </code></pre> <p>Then do the standard rounding thing:</p> <pre><code>roundedrest = (int)(roundedrest + 0.5); </code></pre> <p>And undo the pow.</p> <pre><code>roundednum = pow(roundedrest, -(power)) </code></pre> <p>Where power is the power calculated above.</p> <hr> <p>About accuracy: Pyrolistical's answer is indeed closer to the real result. But note that you can't represent 12.1 exactly in any case. If you print the answers as follows:</p> <pre><code>System.out.println(new BigDecimal(n)); </code></pre> <p>The answers are:</p> <pre><code>Pyro's: 12.0999999999999996447286321199499070644378662109375 Mine: 12.10000000000000142108547152020037174224853515625 Printing 12.1 directly: 12.0999999999999996447286321199499070644378662109375 </code></pre> <p>So, use Pyro's answer!</p>
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    1. PORounding to an arbitrary number of significant digits
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    1. COThis algorithm seems prone to floating point errors. When implemented with JavaScript, I get: 0.06805 -> 0.06810000000000001 and 12.1 -> 12.100000000000001
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    2. CO12.1 by itself can't be represented accurately using floating point - it's not a result of this algorithm.
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    3. COThis code in Java produces 12.100000000000001 and this is using 64-bit doubles which can present 12.1 exactly.
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