Double precision - decimal places

Question

From what I have read, a value of data type double has an approximate precision of 15 decimal places. However, when I use a number whose decimal representation repeats, such as 1.0/7.0, I find that the variable holds the value of 0.14285714285714285 - which is 17 places (via the debugger).

I would like to know why it is represented as 17 places internally, and why a precision of 15 is always written at ~15?

Solution 1

An IEEE double has 53 significant bits (that's the value of DBL_MANT_DIG in <cfloat>). That's approximately 15.95 decimal digits (log10(2⁵³)); the implementation sets DBL_DIG to 15, not 16, because it has to round down. So you have nearly an extra decimal digit of precision (beyond what's implied by DBL_DIG==15) because of that.

The nextafter() function computes the nearest representable number to a given number; it can be used to show just how precise a given number is.

This program:

#include <cstdio>
#include <cfloat>
#include <cmath>

int main() {
    double x = 1.0/7.0;
    printf("FLT_RADIX = %d\n", FLT_RADIX);
    printf("DBL_DIG = %d\n", DBL_DIG);
    printf("DBL_MANT_DIG = %d\n", DBL_MANT_DIG);
    printf("%.17g\n%.17g\n%.17g\n", nextafter(x, 0.0), x, nextafter(x, 1.0));
}

gives me this output on my system:

FLT_RADIX = 2
DBL_DIG = 15
DBL_MANT_DIG = 53
0.14285714285714282
0.14285714285714285
0.14285714285714288

(You can replace %.17g by, say, %.64g to see more digits, none of which are significant.)

As you can see, the last displayed decimal digit changes by 3 with each consecutive value. The fact that the last displayed digit of 1.0/7.0 (5) happens to match the mathematical value is largely coincidental; it was a lucky guess. And the correct rounded digit is 6, not 5. Replacing 1.0/7.0 by 1.0/3.0 gives this output:

FLT_RADIX = 2
DBL_DIG = 15
DBL_MANT_DIG = 53
0.33333333333333326
0.33333333333333331
0.33333333333333337

which shows about 16 decimal digits of precision, as you'd expect.

Solution 2

It is actually 53 binary places, which translates to 15 stable decimal places, meaning that if you round a start out with a number with 15 decimal places, convert it to a double, and then round the double back to 15 decimal places you'll get the same number. To uniquely represent a double you need 17 decimal places (meaning that for every number with 17 decimal places, there's a unique closest double) which is why 17 places are showing up, but not all 17-decimal numbers map to different double values (like in the examples in the other answers).

Solution 3

Decimal representation of floating point numbers is kind of strange. If you have a number with 15 decimal places and convert that to a double, then print it out with exactly 15 decimal places, you should get the same number. On the other hand, if you print out an arbitrary double with 15 decimal places and the convert it back to a double, you won't necessarily get the same value back—you need 17 decimal places for that. And neither 15 nor 17 decimal places are enough to accurately display the exact decimal equivalent of an arbitrary double. In general, you need over 100 decimal places to do that precisely.

See the Wikipedia page for double-precision and this article on floating-point precision.

Solution 4

A double holds 53 binary digits accurately, which is ~15.9545898 decimal digits. The debugger can show as many digits as it pleases to be more accurate to the binary value. Or it might take fewer digits and binary, such as 0.1 takes 1 digit in base 10, but infinite in base 2.

This is odd, so I'll show an extreme example. If we make a super simple floating point value that holds only 3 binary digits of accuracy, and no mantissa or sign (so range is 0-0.875), our options are:

binary - decimal
000    - 0.000
001    - 0.125
010    - 0.250
011    - 0.375
100    - 0.500
101    - 0.625
110    - 0.750
111    - 0.875

But if you do the numbers, this format is only accurate to 0.903089987 decimal digits. Not even 1 digit is accurate. As is easy to see, since there's no value that begins with 0.4?? nor 0.9??, and yet to display the full accuracy, we require 3 decimal digits.

tl;dr: The debugger shows you the value of the floating point variable to some arbitrary precision (19 digits in your case), which doesn't necessarily correlate with the accuracy of the floating point format (17 digits in your case).

Solution 5

IEEE 754 floating point is done in binary. There's no exact conversion from a given number of bits to a given number of decimal digits. 3 bits can hold values from 0 to 7, and 4 bits can hold values from 0 to 15. A value from 0 to 9 takes roughly 3.5 bits, but that's not exact either.

An IEEE 754 double precision number occupies 64 bits. Of this, 52 bits are dedicated to the significand (the rest is a sign bit and exponent). Since the significand is (usually) normalized, there's an implied 53^rd bit.

Now, given 53 bits and roughly 3.5 bits per digit, simple division gives us 15.1429 digits of precision. But remember, that 3.5 bits per decimal digit is only an approximation, not a perfectly accurate answer.

Many (most?) debuggers actually look at the contents of the entire register. On an x86, that's actually an 80-bit number. The x86 floating point unit will normally be adjusted to carry out calculations to 64-bit precision -- but internally, it actually uses a couple of "guard bits", which basically means internally it does the calculation with a few extra bits of precision so it can round the last one correctly. When the debugger looks at the whole register, it'll usually find at least one extra digit that's reasonably accurate -- though since that digit won't have any guard bits, it may not be rounded correctly.

View more solutions