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    <p>You can get most of the effect of rounding to single precision using:</p> <pre><code>y = x + x * 0x1p29 - x * 0x1p29; </code></pre> <p>In most cases, this produces the same result in y as if x had been rounded to float (32-bit binary IEEE 754) and then converted back to double (64-bit). It works by adding a value (x * 0x1p29) that “pushes” some bits of x out of the significand, causing rounding at bit 23, and then subtracting the value that was added. (<code>0x1p29</code> is hexadecimal floating-point for 2<sup>29</sup>, 536870912.)</p> <p>In rare cases, it produces a slightly different result. If you merely want to reduce noise in a model, these rare cases might be negligible. If you want to eliminate them, then, instead of adding and subtracting 2<sup>29</sup>x, you could find the largest power of 2 not greater than x and add and subtract 2<sup>29</sup> times that instead of 2<sup>29</sup>x. (To find the power of 2, you can take the base-two logarithm and take the floor of that. However, there are still rounding issues that might require compensation. Additionally, if the input might be zero or negative, you must avoid the error that occurs when taking its logarithm.)</p> <p>Additionally, this does not reproduce the behavior for numbers that are subnormal in single-precision or that overflow in single-precision.</p> <p>Finally, there are rare cases where computing a double-precision result and then rounding to single precision produces a result slightly different from computing a single-precision result originally, and no method of rounding the double-precision result will fix this.</p>
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    1. POHow to simulate Single precision rounding with Doubles?
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    1. USEric Postpischil
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    1. COWhat's constantly in the back of my mind is that the original system can *consistently* reproduce these values. It's not some random magic going on - it's an algorithm in a computer. At some level the laws of computing say i should be able to reproduce the math.
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      1. USIan Boyd
    2. CO@IanBoyd, you can, see my answer. (ldexp and frexp just twiddle bits, they don't do complicated computations in the background.) So they give you precisely the result you're looking for. (At least if you're using some kind of C variant or Lua. Or, I just checked, Python.)
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    3. CO@IanBoyd: Of course the computation can be replicated. However, there is no reason that should be easy given the operations available in Excel.
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      1. USEric Postpischil
 

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