Precision error in double

Question

A large double value gets changed when printed with %Lf Values upto the following combination gives proper results 9 digits before decimal / 6 digits after decimal e.g. of a value with 9 digits before decimal printed with %Lf Input : 3435537287.32 Output : 3435537287.320000

Once I increase the digits before decimal to 10, the values printed with %Lf adds garbage value. e.g. of a value with 10 digits before decimal printed with %Lf Input : 34355372871.3487 Output : 34355372871.348701 As you can see from the above output the input value is changed.

Is there any compile time option for g++/xlc++ that can be used so that the value is'nt changed?

::Code Snippet::

double d2 =  34355372871.3487; 
double d4 =  3435537287.3487; 

printf("d2 = %lf\n", d2); 
printf("d4 = %lf\n", d4);

Thanks, Hudson

You may find http://floating-point-gui.de/ helpful – Hulk Aug 07 '13 at 12:00 — Hulk, Aug 07 '13 at 12:00

score 4 · Answer 1 · answered Aug 07 '13 at 12:05

4

Short explanation: you need long doubles

long double d2 =  34355372871.3487L; 
long double d4 =  3435537287.3487L; 

printf("d2 = %Lf\n", d2); 
printf("d4 = %Lf\n", d4);

long explanation: your significand takes to much space ;-) Here are the bits used in a Floatingpoint value.

//Type       Sign  Exponent Significand Total bit 
//Double        1        11          52        64 
//Long Double   1        15          64        80

Keep in mind that every digit in decimal takes about 3.3 bits (log(10)/log(2)) so 52bits/3.3 ~ 15-16 digits.

answered Aug 07 '13 at 12:05

Jürgen Schwietering

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Thanks, but even long double gives the same problem. – Hudson Diniz Aug 14 '13 at 13:46
@Hudson: it will if you do not add the L at the end of the constant values. – Jürgen Schwietering Aug 20 '13 at 12:27

score 1 · Answer 2 · answered Aug 07 '13 at 11:57

It is the precision of the double. The double is stored with 64 bits (8bytes).

(http://en.wikipedia.org/wiki/Double-precision_floating-point_format)

It uses 52 bits for a mantissa, 1 bit for the signal and 11 bits for the exponent.

Then, the 52 bits mantissa gives a 15–17 significant decimal digits precision.

score 1 · Answer 3 · answered Aug 07 '13 at 12:16

Due to the limited precision of floating point numbers, not every (or actually, only a few) decimal floating point literals correspond exactly to an existing binary representation.

Every literal is mapped to the closest existing value. This happens for every floating point type, but it will be less noticeable for higher precision types because there are more distinct numbers.

Precision error in double

3 Answers3