C++ – How to print a double value with full precision using cout

Note: the Nintendo 64 does have a 64-bit processor, however:

Many games took advantage of the chip's 32-bit processing mode as the greater data precision available with 64-bit data types is not typically required by 3D games, as well as the fact that processing 64-bit data uses twice as much RAM, cache, and bandwidth, thereby reducing the overall system performance.

From Webopedia:

The term double precision is something of a misnomer because the precision is not really double.
The word double derives from the fact that a double-precision number uses twice as many bits as a regular floating-point number.
For example, if a single-precision number requires 32 bits, its double-precision counterpart will be 64 bits long.

The extra bits increase not only the precision but also the range of magnitudes that can be represented.
The exact amount by which the precision and range of magnitudes are increased depends on what format the program is using to represent floating-point values.
Most computers use a standard format known as the IEEE floating-point format.

The IEEE double-precision format actually has more than twice as many bits of precision as the single-precision format, as well as a much greater range.

From the IEEE standard for floating point arithmetic

Single Precision

The IEEE single precision floating point standard representation requires a 32 bit word, which may be represented as numbered from 0 to 31, left to right.

• The first bit is the sign bit, S,
• the next eight bits are the exponent bits, 'E', and

• the final 23 bits are the fraction 'F':

  S EEEEEEEE FFFFFFFFFFFFFFFFFFFFFFF
  0 1      8 9                    31
  

The value V represented by the word may be determined as follows:

• If E=255 and F is nonzero, then V=NaN ("Not a number")
• If E=255 and F is zero and S is 1, then V=-Infinity
• If E=255 and F is zero and S is 0, then V=Infinity
• If 0<E<255 then V=(-1)**S * 2 ** (E-127) * (1.F) where "1.F" is intended to represent the binary number created by prefixing F with an implicit leading 1 and a binary point.
• If E=0 and F is nonzero, then V=(-1)**S * 2 ** (-126) * (0.F). These are "unnormalized" values.
• If E=0 and F is zero and S is 1, then V=-0
• If E=0 and F is zero and S is 0, then V=0

In particular,

0 00000000 00000000000000000000000 = 0
1 00000000 00000000000000000000000 = -0

0 11111111 00000000000000000000000 = Infinity
1 11111111 00000000000000000000000 = -Infinity

0 11111111 00000100000000000000000 = NaN
1 11111111 00100010001001010101010 = NaN

0 10000000 00000000000000000000000 = +1 * 2**(128-127) * 1.0 = 2
0 10000001 10100000000000000000000 = +1 * 2**(129-127) * 1.101 = 6.5
1 10000001 10100000000000000000000 = -1 * 2**(129-127) * 1.101 = -6.5

0 00000001 00000000000000000000000 = +1 * 2**(1-127) * 1.0 = 2**(-126)
0 00000000 10000000000000000000000 = +1 * 2**(-126) * 0.1 = 2**(-127) 
0 00000000 00000000000000000000001 = +1 * 2**(-126) * 
                                     0.00000000000000000000001 = 
                                     2**(-149)  (Smallest positive value)

Double Precision

The IEEE double precision floating point standard representation requires a 64 bit word, which may be represented as numbered from 0 to 63, left to right.

• The first bit is the sign bit, S,
• the next eleven bits are the exponent bits, 'E', and

• the final 52 bits are the fraction 'F':

  S EEEEEEEEEEE FFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFF
  0 1        11 12                                                63
  

The value V represented by the word may be determined as follows:

• If E=2047 and F is nonzero, then V=NaN ("Not a number")
• If E=2047 and F is zero and S is 1, then V=-Infinity
• If E=2047 and F is zero and S is 0, then V=Infinity
• If 0<E<2047 then V=(-1)**S * 2 ** (E-1023) * (1.F) where "1.F" is intended to represent the binary number created by prefixing F with an implicit leading 1 and a binary point.
• If E=0 and F is nonzero, then V=(-1)**S * 2 ** (-1022) * (0.F) These are "unnormalized" values.
• If E=0 and F is zero and S is 1, then V=-0
• If E=0 and F is zero and S is 0, then V=0

Reference:
ANSI/IEEE Standard 754-1985,
Standard for Binary Floating Point Arithmetic.

From the Floating-Point Guide:

What can I do to avoid this problem?

That depends on what kind of calculations you’re doing.

• If you really need your results to add up exactly, especially when you work with money: use a special decimal datatype.
• If you just don’t want to see all those extra decimal places: simply format your result rounded to a fixed number of decimal places when displaying it.
• If you have no decimal datatype available, an alternative is to work with integers, e.g. do money calculations entirely in cents. But this is more work and has some drawbacks.

Note that the first point only applies if you really need specific precise decimal behaviour. Most people don't need that, they're just irritated that their programs don't work correctly with numbers like 1/10 without realizing that they wouldn't even blink at the same error if it occurred with 1/3.

If the first point really applies to you, use BigDecimal for JavaScript, which is not elegant at all, but actually solves the problem rather than providing an imperfect workaround.