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    <p>The reason that the extra decimal places appear is likely because of a slight precision loss when converting from double to float (floating point conversions are never perfect).</p> <p>If you need to round the number then after you've done the conversion you can do:</p> <pre><code>result = Math.Round(result, 2); </code></pre> <p>Edit: Given Eric's comment below, here is some extra info:</p> <p>A <code>float</code> in C# is a 32 bit floating point representation of a number, whereas a <code>double</code> is 64 bit. Not every number is perfectly representable by a floating point number. When you convert from the 32 bit float to the 64 bit double you've got 32 extra bits to fill in with information. This extra information ends up being the extra digits that you see:</p> <pre><code>float foo = 50.8467178f; double bar = System.Convert.ToDouble(foo); System.Console.WriteLine(bar); // prints 50.8467178344727 on my machine </code></pre> <p>Since the original number is not perfectly representable in floating point, the output ends up being a slightly different number because of the conversion process.</p> <p>If you need to maintain precision for this number you'll need to use <code>decimal</code> values instead:</p> <pre><code>decimal foo = 50.8467178M; double bar = System.Convert.ToDouble(foo); System.Console.WriteLine(bar); // prints 50.8467178 on my machine </code></pre> <p>If you can't use <code>decimal</code> (e.g. the <code>float</code> is the output of a third party library or something) then the rounding approach is the next best option:</p> <pre><code>float foo = 50.8467178f; double bar = System.Convert.ToDouble(foo); bar = System.Math.Round(bar, 7); System.Console.WriteLine(bar); // prints 50.8467178 on my machine </code></pre>
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    1. PORounding numbers in C# producing too many decimal places
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    1. USAdam Rodger
    1. USAdam Rodger
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    1. COThis answer is correct that part of the problem is the rounding error caused by conversion to float, but this is incomplete because it does not explain why displaying a value derived from a narrower precision produces more digits than displaying a value derived from a wider precision. It is also incorrect to state that floating-point conversions are never perfect. In IEEE-754, conversions from `float` to `double` are error free, and conversions from `double` to `float` are error free if the original value is representable as a `float`.
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      1. USEric Postpischil
    2. CO@EricPostpischil: Although IEEE specifies that an operation x+y must yield the best floating-point representation of the nominal value of x plus the nominal value of y, that does not mean that x really "represents" its nominal value. The nominal values of 0.1f and (double)0.1f may be the same, but the semantic meanings differ. There is no double value whose semantic meaning matches that of 0.1f. Indeed, I would posit that converting a float to a double generally alters its semantic meaning more than does converting a double to a float.
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      1. USsupercat
    3. CO@supercat: (a) This notion of “semantic meaning” is outside the bounds of the programming language and not relevant to this question. (b) By this standard, the semantics meanings of `2` and `2l` differ, so integer conversions are never perfect.
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      1. USEric Postpischil
 

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