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POHow can I check whether two numbers are within "x" significant figures of the precision limits of the floating point type?
 

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  1. POHow can I check whether two numbers are within "x" significant figures of the precision limits of the floating point type?
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    <p>So suppose we have a float type XType in which we have two numbers:</p> <pre><code>XType const a = 1.2345 XType const b = 1.2300 </code></pre> <p>Then I want a function IsClose(XType const f1,XType const f2,unsigned const truncated_figures) such that</p> <pre><code>// the numbers are equal if the last two figures are ignored (1.23 == 1.23) IsClose&lt;XType&gt;(a,b,2) == true // the numbers are not equal if only the last is ignored (1.234 != 1.230) IsClose&lt;XType&gt;(a,b,1) == false </code></pre> <p>So far I have this ugly mess, but I'm yet to convince myself it's correct:</p> <pre><code>// check if two floating point numbers are close to within "figures_tolerance" figures of precision for the applicable type template &lt;typename FloatType&gt; bool const IsClose(FloatType const f1, FloatType const f2, unsigned const figures_tolerance) { FloatType const tolerance_exponent = std::pow(10.0,figures_tolerance); FloatType const tolerance = std::pow(tolerance_exponent,std::log10(f1)) * std::numeric_limits&lt;FloatType&gt;::epsilon() ; return std::abs(f1 - f2) &lt; tolerance; } </code></pre> <p>My reasoning is that the tolerance should be the epsilon raised to the order of magnitude that the number exceeds or subseeds 1.0 (the significant figures for which the epsilon is based). Does this make sense? Is there a better, more reliable way?</p> <p>EDIT: My solution using the template function is below (it is based on user763305's answer below)</p> <pre><code>// check if two floating point numbers are within the last n digits of precision for the // largest of the two numbers being compared. template &lt;typename FloatType&gt; bool const IsWithinPrecision(FloatType const f1, FloatType const f2, unsigned const n = 1U) { FloatType const f_ref = std::max(std::abs(f1), std::abs(f2)); FloatType const distance = std::abs(f1 - f2); FloatType const e = std::numeric_limits&lt;FloatType&gt;::epsilon(); return distance &lt; std::pow((FloatType) 10.0, (FloatType) n) * e * f_ref; } </code></pre>
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      1. POHow can I check whether two numbers are within "x" significant figures of the precision limits of the floating point type?
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      1. POHow can I check whether two numbers are within "x" significant figures of the precision limits of the floating point type?
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    1. COWhat are you trying to do? Your question conflates different things. If you are trying to estimate how much error there could be in floating-point types, then it is impossible to do it with a formula, because the error depends on the specific operations and values used. It varies greatly between different kinds of computations and can range from zero to infinity.
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      1. POHow can I check whether two numbers are within "x" significant figures of the precision limits of the floating point type?
      1. USEric Postpischil
    2. COOn the other hand, if you are trying to determine whether two numbers are the same in the first `x` significant digits, then this code is wrong. The epsilon of the floating-point type would not be much involved. You would determine the first `x` digits of each and then compare them.
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      1. POHow can I check whether two numbers are within "x" significant figures of the precision limits of the floating point type?
      1. USEric Postpischil
    3. CO`IsClose` sounds a little subjective, I would assume `IsClose(1.2345,1.2300)` is close no matter the figure tolerance. But wait, that also depends on the context its used in, if the little precision is a big matter, then `IsClose` between the two isn't close. In what context will you use this mostly?
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