'float' vs. 'double' precision

The code

float x = 3.141592653589793238;
double z = 3.141592653589793238;
printf("x=%f\n", x);
printf("z=%f\n", z);
printf("x=%20.18f\n", x);
printf("z=%20.18f\n", z);

will give you the output

x=3.141593
z=3.141593
x=3.141592741012573242
z=3.141592653589793116

where on the third line of output `741012573242` is garbage and on the fourth line `116` is garbage. Do doubles always have 16 significant figures while floats always have 7 significant figures? Why don't doubles have 14 significant figures?

It's not exactly *double* precision because of how [IEEE 754](

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) works, and because binary doesn't really translate well to decimal. Take a look at the standard if you're interested.

It's usually based on significant figures of both the exponent and significand in base 2, not base 10. From what I can tell in the C99 standard, however, there is no specified precision for floats and doubles (other than the fact that 1 and `1 + 1E-5` / `1 + 1E-7` are distinguishable [`float` and `double` repsectively]). However, the number of significant figures is left to the implementer (as well as which base they use internally, so in other words, an implementation could decide to make it based on 18 digits of precision in base 3). [1]

If you need to know these values, the constants `FLT_RADIX` and `FLT_MANT_DIG` (and `DBL_MANT_DIG` / `LDBL_MANT_DIG`) are defined in float.h.

The reason it's called a `double` is because the number of bytes used to store it is double the number of a float (but this includes both the exponent and significand). The IEEE 754 standard (used by most compilers) allocate relatively more bits for the significand than the exponent (23 to 9 for `float` vs. 52 to 12 for `double`), which is why the precision is more than doubled.

1: Section 5.2.4.2.2 ( )

> Do doubles always have 16 significant
> figures while floats always have 7
> significant figures?

No. Doubles always have 53 significant **bits** and floats always have 24 significant **bits** (except for denormals, infinities, and NaN values, but those are subjects for a different question). These are binary formats, and you can only speak clearly about the precision of their representations in terms of binary digits (bits).

This is analogous to the question of how many digits can be stored in a binary integer: an unsigned 32 bit integer can store integers with up to 32 bits, which doesn't precisely map to any number of decimal digits: all integers of up to 9 decimal digits can be stored, but a lot of 10-digit numbers can be stored as well.

> Why don't doubles
> have 14 significant figures?

The encoding of a double uses 64 bits (1 bit for the sign, 11 bits for the exponent, 52 explicit significant bits and one implicit bit), which is *double* the number of bits used to represent a float (32 bits).

Floating point numbers in C use [IEEE 754][1] encoding.

This type of encoding uses a sign, a significand, and an exponent.

Because of this encoding, many numbers will have small changes to allow them to be stored.

Also, the number of significant digits can change slightly since it is a binary representation, not a decimal one.

Single precision (float) gives you 23 bits of significand, 8 bits of exponent, and 1 sign bit.

Double precision (double) gives you 52 bits of significand, 11 bits of exponent, and 1 sign bit.


[1]:

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