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copied!<p>A double has the precision of 52 binary digits or 15.95 decimal digits. See <a href="http://en.wikipedia.org/wiki/IEEE_754-2008" rel="nofollow">http://en.wikipedia.org/wiki/IEEE_754-2008</a>. You need at least 16 decimal digits to record the full precision of a double in all cases. [But see fourth edit, below].</p> <p>By the way, this means significant digits.</p> <p>Answer to OP edits:</p> <p>Your floating point to decimal string runtime is outputing way more digits than are significant. A double can only hold 52 bits of significand (actually, 53, if you count a "hidden" 1 that is not stored). That means the the resolution is not more than 2 ^ -53 = 1.11e-16.</p> <p>For example: 1 + 2 ^ -52 = 1.0000000000000002220446049250313 . . . .</p> <p>Those decimal digits, .0000000000000002220446049250313 . . . . are the smallest binary "step" in a double <strong>when converted to decimal.</strong></p> <p>The "step" inside the double is:</p> <p>.0000000000000000000000000000000000000000000000000001 <strong>in binary.</strong></p> <p>Note that the binary step is exact, while the decimal step is inexact.</p> <p>Hence the decimal representation above,</p> <p>1.0000000000000002220446049250313 . . . </p> <p>is an inexact representation of the exact binary number: </p> <p>1.0000000000000000000000000000000000000000000000000001.</p> <p><strong>Third Edit:</strong></p> <p>The next possible value for a double, which in exact binary is:</p> <p>1.0000000000000000000000000000000000000000000000000010</p> <p>converts inexactly in decimal to </p> <p>1.0000000000000004440892098500626 . . . .</p> <p>So all of those extra digits in the decimal are not really significant, they are just <em>base conversion artifacts.</em></p> <p><strong>Fourth Edit:</strong></p> <p>Though a double stores at most 16 significant decimal digits, <strong>sometimes 17 decimal digits are necessary to represent the number</strong>. The reason has to do with <em>digit slicing</em>.</p> <p>As I mentioned above, there are 52 + 1 binary digits in the double. The "+ 1" is an assumed leading 1, and is neither stored nor significant. In the case of an integer, those 52 binary digits form a number between 0 and 2^53 - 1. How many decimal digits are necessary to store such a number? Well, log_10 (2^53 - 1) is about 15.95. So at most 16 decimal digits are necessary. Let's label these d_0 to d_15.</p> <p>Now consider that IEEE floating point numbers also have an <em>binary</em> exponent. What happens when we increment the exponet by, say, 2? We have multiplied our 52-bit number, whatever it was, by 4. Now, instead of our 52 binary digits aligning perfectly with our decimal digits d_0 to d_15, we have some significant binary digits represented in d_16. However, since we multiplied by something less than 10, we still have significant binary digits represented in d_0. So our 15.95 decimal digits now occuply d_1 to d_15, plus some upper bits of d_0 and some lower bits of d_16. This is why 17 decimal digits are sometimes needed to represent a IEEE double.</p> <p><strong>Fifth Edit</strong></p> <p>Fixed numerical errors</p>

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