StackOverflow2013

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2013-04-25T15:27:36.957
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<blockquote> After that, however, it subtracted (dividend/2^31) from the result. </blockquote> Actually, it subtracts <code>dividend >> 31</code>, which is <code>-1</code> for negative <code>dividend</code>, and 0 for non-negative dividend, when right-shifting negative integers is arithmetic right-shift (and <code>int</code> is 32 bits wide). <pre><code>0x6666667 = (2^34 + 6)/10 </code></pre> So for <code>x < 0</code>, we have, writing <code>x = 10*k + r</code> with <code>-10 < r <= 0</code>, <pre><code>0x66666667 * (10*k+r) = (2^34+6)*k + (2^34 + 6)*r/10 = 2^34*k + 6*k + (2^34+6)*r/10 </code></pre> Now, arithmetic right shift of negative integers yields the floor of <code>v / 2^n</code>, so <pre><code>(0x66666667 * x) >> 34 </code></pre> results in <pre><code>k + floor((6*k + (2^34+6)*r/10) / 2^34) </code></pre> So we need to see that <pre><code>-2^34 < 6*k + (2^34+6)*r/10 < 0 </code></pre> The right inequality is easy, both <code>k</code> and <code>r</code> are non-positive, and not both are 0. For the left inequality, a bit more analysis is needed. <pre><code>r >= -9 </code></pre> so the absolute value of <code>(2^34+6)*r/10</code> is at most <code>2^34+6 - (2^34+6)/10</code>. <pre><code>|k| <= 2^31/10, </code></pre> so <code>|6*k| <= 3*2^31/5</code>. And it remains to verify that <pre><code>6 + 3*2^31/5 < (2^34+6)/10 1288490194 < 1717986919 </code></pre> Yup, true.
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1. This table or related slice is empty.
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1. POFast Division on GCC/ARM
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1. PTAnswer
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1. USDaniel Fischer
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1. POFast Division on GCC/ARM
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1. VO
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2. VO
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1. COThanks for the elaboration.
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 PO
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 USDaniel Scocco

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