What's the difference between a single precision and double precision floating point operation?

Question

What is the difference between a single precision floating point operation and double precision floating operation?

I\'m especially interested in practical terms in

Answer 1

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.

Answer 2

Single precision number uses 32 bits, with the MSB being sign bit, whereas double precision number uses 64bits, MSB being sign bit

Single precision

SEEEEEEEEFFFFFFFFFFFFFFFFFFFFFFF.(SIGN+EXPONENT+SIGNIFICAND)

Double precision:

SEEEEEEEEEEEFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFFF.(SIGN+EXPONENT+SIGNIFICAND)

Answer 3

As to the question "Can the ps3 and xbxo 360 pull off double precision floating point operations or only single precision and in generel use is the double precision capabilities made use of (if they exist?)."

I believe that both platforms are incapable of double floating point. The original Cell processor only had 32 bit floats, same with the ATI hardware which the XBox 360 is based on (R600). The Cell got double floating point support later on, but I'm pretty sure the PS3 doesn't use that chippery.

Answer 4

Basically single precision floating point arithmetic deals with 32 bit floating point numbers whereas double precision deals with 64 bit.

The number of bits in double precision increases the maximum value that can be stored as well as increasing the precision (ie the number of significant digits).

Answer 5

Okay, the basic difference at the machine is that double precision uses twice as many bits as single. In the usual implementation,that's 32 bits for single, 64 bits for double.

But what does that mean? If we assume the IEEE standard, then a single precision number has about 23 bits of the mantissa, and a maximum exponent of about 38; a double precision has 52 bits for the mantissa, and a maximum exponent of about 308.

The details are at Wikipedia, as usual.