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    <p>Because <code>0.1</code> isn't 0.1; that value isn't representable in double precision, so it gets rounded to the nearest double-precision number, which is exactly:</p> <pre><code>0.1000000000000000055511151231257827021181583404541015625 </code></pre> <p>When you call <code>fmod</code>, you get the remainder of division by the value listed above, which is exactly:</p> <pre><code>0.0999999999999999500399638918679556809365749359130859375 </code></pre> <p>which rounds to <code>0.1</code> (or maybe <code>0.09999999999999995</code>) when you print it.</p> <p>In other words, <code>fmod</code> works perfectly, but you're not giving it the input that you think you are.</p> <hr> <p><strong>Edit:</strong> Your own implementation gives you the correct answer because it is <em>less accurate</em>, believe it or not. First off, note that <code>fmod</code> computes the remainder without any rounding error; the only source of inaccuracy is the representation error introduced by using the value <code>0.1</code>. Now, let's walk through your implementation, and see how the rounding error that it incurs exactly cancels out the representation error.</p> <p>Evaluate <code>a - floor(a/n) * n</code> one step at a time, keeping track of the exact values computed at each stage:</p> <p>First we evaluate <code>1.0/n</code>, where <code>n</code> is the closest double-precision approximation to <code>0.1</code> as shown above. The result of this division is approximately:</p> <pre><code>9.999999999999999444888487687421760603063276150363492645647081359... </code></pre> <p>Note that this value is not a representable double precision number -- so it gets <em>rounded</em>. To see how this rounding happens, let's look at the number in binary instead of decimal:</p> <pre><code>1001.1111111111111111111111111111111111111111111111111 10110000000... </code></pre> <p>The space indicates where the rounding to double precision occurs. Since the part after the round point is larger than the exact half-way point, this value rounds up to exactly <code>10</code>.</p> <p><code>floor(10.0)</code> is, predictably, <code>10.0</code>. So all that's left is to compute <code>1.0 - 10.0*0.1</code>.</p> <p>In binary, the exact value of <code>10.0 * 0.1</code> is:</p> <pre><code>1.0000000000000000000000000000000000000000000000000000 0100 </code></pre> <p>again, this value is not representable as a double, and so is rounded at the position indicated by a space. This time it rounds down to exactly <code>1.0</code>, and so the final computation is <code>1.0 - 1.0</code>, which is of course <code>0.0</code>.</p> <p>Your implementation contains two rounding errors, which happen to exactly cancel out the representation error of the value <code>0.1</code> in this case. <code>fmod</code>, by contrast, is <em>always</em> exact (at least on platforms with a good numerics library), and exposes the representation error of <code>0.1</code>.</p>
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    1. POWhy fmod(1.0,0.1) == .1?
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    1. COThis doesn't explain why `1.0 - floor(1.0/0.1) * 0.1` gives zero though.
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    2. COyeah, I get it. 1.0 can be exactly represented by a double while 0.1 can't. Since the closest representable number is greater than the original the result of the division will be something like 9.99999999999, which becomes 9 after truncation.
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    3. CO@KennyTM: added an answer to the follow-up question for you.
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