What is the minimum number of significant decimal digits in a floating point literal to represent the value as correct as possible?

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For example, using IEEE-754 32-bit binary floating points, let's represent the value of 1 / 3. It cannot be done exactly, but 0x3eaaaaab produces the closest value to 1 / 3. You might want to write the value in decimal, and let the compiler to convert the decimal literal to a binary floating point number.

0.333333f    -> 0x3eaaaa9f (0.333332986)
0.3333333f   -> 0x3eaaaaaa (0.333333313)
0.33333333f  -> 0x3eaaaaab (0.333333343)
0.333333333f -> 0x3eaaaaab (0.333333343)

You can see that 8 (significant) decimal digits is enough to represent the value as correct as possible (closest to the actual value).

I tested with π and e (base of the natural log), and both needed 8 decimal digits for the correctest.

3.14159f    -> 0x40490fd0 (3.14159012)
3.141593f   -> 0x40490fdc (3.14159298)
3.1415927f  -> 0x40490fdb (3.14159274)
3.14159265f -> 0x40490fdb (3.14159274)

2.71828f    -> 0x402df84d (2.71828008)
2.718282f   -> 0x402df855 (2.71828198)
2.7182818f  -> 0x402df854 (2.71828175)
2.71828183f -> 0x402df854 (2.71828175)

However, √2 appears to need 9 digits.

1.41421f     -> 0x3fb504d5 (1.41420996)
1.414214f    -> 0x3fb504f7 (1.41421402)
1.4142136f   -> 0x3fb504f4 (1.41421366)
1.41421356f  -> 0x3fb504f3 (1.41421354)
1.414213562f -> 0x3fb504f3 (1.41421354)

https://godbolt.org/z/W5vEcs695

Looking at these results, it's probably right that a decimal floating-point literal with 9 significant digits is sufficient to produce a most correct 32-bit binary floating point value, and in practice something like 12~15 digits would work for sure if space for storing the extra digits doesn't matter.

But I'm interested in the math behind it. How can one be sure that 9 digits is enough in this case? What about double or even arbitrary precision, is there a simple formula to derive the number of digits needed?

The current answers and the links in the comments confirm that 9 digits is enough for most cases, but I've found a counterexample where 9 digits is not enough. In fact, infinite precision in the decimal format is required to be always correctly converted (rounded to the closest) to some binary floating point format (IEEE-754 binary32 floats for the discussion).

8388609.499 represented with 9 significant decimal digits is 8388609.50. This number converted to float has the value of 8388610. On the other hand, the number represented with 10 or more digits will always preserve the original value, and this number converted to float has the value 8388609.

You can see 8388609.499 needs more than 9 digits to be most accurately converted to float. There are infinitely many such numbers, placed very close to the half point of two representable values in the binary float format.

numerical-methods ieee-754 precision floating-point c
3 Answers

I think you are looking for *_DECIMAL_DIG constants. C standard provides small explanation and formula on how they are calculated (N2176 C17 draft):

5.2.4.2.2 Characteristics of floating types <float.h>

  1. The values given in the following list shall be replaced by constant expressions with implementation-defined values that are greater or equal in magnitude (absolute value) to those shown, with the same sign:

    ...

    • number of decimal digits, n, such that any floating-point number with p radix b digits can be rounded to a floating-point number with n decimal digits and back again without change to the value,

      p log10 b        if b is a power of 10
⌈1 + p log10 b⌉  otherwise


FLT_DECIMAL_DIG  6
DBL_DECIMAL_DIG  10
LDBL_DECIMAL_DIG 10

With IEEE-754 32-bit float b = FLT_RADIX = 2 and p = FLT_MANT_DIG = 24, result is FLT_DECIMAL_DIG = ⌈1 + 24 log10 2⌉ = 9. (⌈x⌉=ceil(x)) is ceiling function: round result up)

user_6947330

What about double or even arbitrary precision, is there a simple formula to derive the number of digits needed?>

From C17 § 5.2.4.2.2 11 FLT_DECIMAL_DIG, DBL_DECIMAL_DIG, LDBL_DECIMAL_DIG

number of decimal digits, n, such that any floating-point number with p radix b digits can be rounded to a floating-point number with n decimal digits and back again without change to the value,

pmax log10 b: if b is a power of 10
1 + pmax log10 b: otherwise

But I'm interested in the math behind it. How can one be sure that 9 digits is enough in this case?

Each range of binary floating point like [1.0 ... 2.0), [128.0 ... 256.0), [0.125 ... 0.5) contains 2p - 1 values uniformly distributed. e.g. With float, p = 24.

Each range of a decade of decimal text with n significant digits in exponential notation like [1.0 ... 9.999...), [100.0f ... 999.999...), [0.001 ... 0.00999...) contains 10n - 1 values uniformly distributed.

Example: common float:
When p is 24 with 224 combinations, n must at least 8 to form the 16,777,216 combinations to distinctly round-trip float to decimal text to float. As the end-points of two decimal ranges above may exist well within that set of 224, the larger decimal values are spaced out further apart. This necessitates a +1 decimal digit.

Example:

Consider the 2 adjacent float values

10.000009_5367431640625
10.000010_49041748046875

Both convert to 8 significant digits decimal text "10.000010". 8 is not enough.

9 is always enough as we do not need more than 167,772,160 to distinguish 16,777,216 float values.

OP also asks about 8388609.499. (Let us only consider float for simplicity.)

That value is nearly half-way between 2 float values.

8388609.0f  // Nearest lower float value
8388609.499 // OP's constant as code
8388610.0f  // Nearest upper float value

OP reports: "You can see 8388609.499 needs more than 9 digits to be most accurately converted to float."

And let us review the title "What is the minimum number of significant decimal digits in a floating point literal*1 to represent the value as correct as possible?"

This new question part emphasizes that the value in question is the value of the source code 8388609.499 and not the floating point constant it becomes in emitted code: 8388608.0f.

If we consider the value to be the value of the floating point constant, only up to 9 significant decimal digits are needed to define the floating point constant 8388608.0f. 8388608.49, as source code is sufficient.

But to get the closest floating point constant based on some number as code yes indeed could take many digits.

Consider the typical smallest float, FLT_TRUE_MIN with the exact decimal value of :

0.00000000000000000000000000000000000000000000140129846432481707092372958328991613128026194187651577175706828388979108268586060148663818836212158203125

Half way between that and 0.0 is 0.000..(~39 more zeroes)..0007006..(~ 100 more digits)..15625.

It that last digit was 6 or 4, the closest float would be FLT_TRUE_MIN or 0.0f respectively. So now we have a case where 109 significant digits are "needed" to select between 2 possible float.

To forego us going over the cliffs of insanity, IEEE-758 has already addressed this.

The number of significant decimal digits a translation (compiler) must examine to be compliant with that spec (not necessarily the C spec) is far more limited, even if the extra digits could translate to another FP value.

IIRC, it is in effect FLT_DECIMAL_DIG + 3. So for a common float, as little as 9 + 3 significant decimal digits may be examined.

[Edit]

correct rounding is only guaranteed for the number of decimal digits required plus 3 for the largest supported binary format.

*1 C does not define: floating point literal, but does define floating point constant, so that term is used.

user_24103590

What is the minimum number of significant decimal digits in a floating point literal to represent the value as correct as possible?

There is no guarantee from the C standard that any number of decimal digits in a floating-point literal will produce the nearest value actually representable in the floating-point format. In discussing floating-point literals, C 2018 6.4.4.2 3 says:

… For decimal floating constants, … the result is either the nearest representable value, or the larger or smaller representable value immediately adjacent to the nearest representable value, chosen in an implementation-defined manner…

For quality, C implementations should correctly round floating-point literals to the nearest representable value, with ties going to the choice with the even low digit. In that case, the FLT_DECIMAL_DIG, DBL_DECIMAL_DIG, and LDBL_DECIMAL_DIG values defined in <float.h> provide numbers of digits that always suffice to uniquely identify a representable value.

How can one be sure that 9 digits is enough in this case?

You need statements to this effect in the compiler documentation, such as statements that it provides correct rounding for floating-point literals and that it uses IEEE-754 binary32 (a.k.a. “single precision”) for float (or some other format that would only require nine significant digits to uniquely identify all representable values).

What about double or even arbitrary precision, is there a simple formula to derive the number of digits needed?

The C standard indicates the constants above are calculated as p log10 b if b is a power of ten and ceil(1 + p log10 b) otherwise, where p is the number of digits in the floating-point format and b is the base used in the format. These always suffice, but the latter is not always necessary. The latter provides the number of digits needed if the exponent range were unbounded; its “1 +” covers all possible allowances for how the powers of b interact with the powers of 10, in a sense. But any floating-point format has a finite exponent range, and, for some choices of exponent range, ceil(p log10 b) would suffice instead of ceil(1 + p log10 b). There is no simple formula for this. It does not occur with the standard IEEE-754 formats and can be neglected in practice.

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