CODEWITHSUNDEEP
c++floating-pointprecision
markzzz
markzzz

Reputation: 47995

Shouldn't Float 6 and Double 15 available digits?

I used to think that float can use max 6 digit and double 15, after comma. But if I print limits here:

typedef std::numeric_limits<float> fl;
typedef std::numeric_limits<double> dbl;

int main()
{
    std::cout << fl::max_digits10 << std::endl;
    std::cout << dbl::max_digits10 << std::endl;
}

It prints float 9 and double 17?

Upvotes: 1

Views: 247

Answers (3)

Paul Floyd
Paul Floyd

Reputation: 6936

There isn't an exact correspondence between decimal digits and binary digits.

IEEE 754 single precision uses 23 bits plus 1 for the implicit leading 1. Double precision uses 52+1 bits.

To get the equivalent decimal precision, use

log10(2^binary_digits) = binary_digits*log10(2)

For single precision this is

24*log10(2) = 7.22

and for double precision

53*log10(2) = 15.95

See here and also the Wikipedia page which I don't find to be particularly concise.

Upvotes: 0

Amadeus
Amadeus

Reputation: 10665

From cppreference:

Unlike most mathematical operations, the conversion of a floating-point value to text and back is exact as long as at least max_digits10 were used (9 for float, 17 for double): it is guaranteed to produce the same floating-point value, even though the intermediate text representation is not exact. It may take over a hundred decimal digits to represent the precise value of a float in decimal notation.

For example (I am using http://www.exploringbinary.com/floating-point-converter/ to facilitate the conversion) and double as the precision format:

1.1e308 => 109999999999999997216016380169010472601796114571365898835589230322558260940308155816455878138416026219051443651421887588487855623732463609216261733330773329156055234383563489264255892767376061912596780024055526930962873899746391708729279405123637426157351830292874541601579169431016577315555383826285225574400

Using 16 significant digits:

1.099999999999999e308 => 109999999999999897424000903433019889783160462729437595463026208549681185812946033955861284690212736971153169019636833121365513414107701410594362313651090292197465320141992473263972245213092236035710707805906167798295036672550192042188756649080117981714588407890666666245533825643214495197630622309084729180160

Using 17 significant digits:

1.0999999999999999e308 => 109999999999999997216016380169010472601796114571365898835589230322558260940308155816455878138416026219051443651421887588487855623732463609216261733330773329156055234383563489264255892767376061912596780024055526930962873899746391708729279405123637426157351830292874541601579169431016577315555383826285225574400

which is the same as the original

More than 17 significant digits:

1.09999999999999995555e308 => 109999999999999997216016380169010472601796114571365898835589230322558260940308155816455878138416026219051443651421887588487855623732463609216261733330773329156055234383563489264255892767376061912596780024055526930962873899746391708729279405123637426157351830292874541601579169431016577315555383826285225574400

Continue to be the same as the original.

Upvotes: 2

user743382
user743382

Reputation:

You're confusing digits10 and max_digits10.

If digits10 is 6, then any number with six decimal digits can be converted to the floating point type, and back, and when rounded back to six decimal digits, produces the original value.

If max_digits10 is 9, then there exist at least two floating point numbers that when converted to decimal produce the same initial 8 decimal digits.

digits10 is the number you're looking for, based on your description. It's about converting from decimal to binary floating point back to decimal.

max_digits10 is a number about converting from binary floating point to decimal back to binary floating point.

Upvotes: 3

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