Article Q42980
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Tutorial to Understand IEEE Floating-Point Errors - Q42980
Floating-point mathematics is a complex topic that confuses many
programmers. The tutorial below should help you recognize programming
situations where floating-point errors are likely to occur and how to
avoid them. It should also allow you to recognize cases that are
caused by inherent floating-point math limitations as opposed to
actual compiler bugs.
The information in this article applies to:
- The Standard and Professional Editions of Microsoft Visual Basic
version 1.0 for MS-DOS
- Microsoft Visual Basic version 1.0 for Windows
- Microsoft QuickBasic versions 4.0, 4.0b, and 4.5 for MS-DOS
- Microsoft Basic Professional Development System (PDS) versions 7.0
and 7.1 for MS-DOS
- Microsoft Basic Compiler versions 6.0 and 6.0b for MS-DOS
More Information:
Decimal and Binary Number Systems
---------------------------------
Normally, we count things in base 10. The base is completely
arbitrary. (The only reason that people have traditionally used base
10 is that they have 10 fingers, which have made handy counting
tools).
The number 532.25 in decimal (base 10) means the following:
(5 * 10^2) + (3 * 10^1) + (2 * 10^0) + (2 * 10^-1) + (5 * 10^-2)
500 + 30 + 2 + 2/10 + 5/100
_________
= 532.25
In the binary number system (base 2), each column represents a power
of 2 instead of 10. For example, the number 101.01 means the
following:
(1 * 2^2) + (0 * 2^1) + (1 * 2^0) + (0 * 2^-1) + (1 * 2^-2)
4 + 0 + 1 + 0 + 1/4
_________
= 5.25 Decimal
How Integers Are Represented in PCs
-----------------------------------
Since there is no fractional part to an integer, its machine
representation is much simpler than it is for floating-point values.
Normal integers on PCs are 2 bytes (16 bits) long with the most
significant bit indicating the sign. Long integers are 4 bytes long.
Positive values are straightforward binary numbers. For example:
1 Decimal = 1 Binary
2 Decimal = 10 Binary
22 Decimal = 10110 Binary etc.
However, negative integers are represented using the two's complement
scheme. To get the two's complement representation for a negative
number, take the binary representation for the number's absolute value
and then flip all the bits and add 1. For example:
4 Decimal = 0000 0000 0000 0100
1111 1111 1111 1011 Flip the Bits
-4 = 1111 1111 1111 1100 Add 1
Note that -1 Decimal = 1111 1111 1111 1111 in Binary, which explains
why BASIC treats -1 as logical true (All bits = 1). This is a
consequence of not having separate operators for bitwise and logical
comparisons. Often in BASIC, it is convenient to use the code fragment
below when your program will be making many logical comparisons. This
greatly aids readability.
CONST TRUE = -1
CONST FALSE = NOT TRUE
Note that adding any combination of two's complement numbers together
using ordinary binary arithmetic produces the correct result.
Floating-Point Complications
----------------------------
Every decimal integer can be exactly represented by a binary integer;
however, this is not true for fractional numbers. In fact, every
number that is irrational in base 10 will also be irrational in any
system with a base smaller than 10.
For binary, in particular, only fractional numbers that can be
represented in the form p/q, where q is an integer power of 2, can be
expressed exactly, with a finite number of bits.
Even common decimal fractions, such as decimal 0.0001, cannot be
represented exactly in binary. (0.0001 is a repeating binary fraction
with a period of 104 bits!.)
This explains why a simple example, such as the following
SUM = 0
FOR I% = 1 TO 10000
SUM = SUM + 0.0001
NEXT I%
PRINT SUM ' Theoretically = 1.0.
will PRINT 1.000054 as output. The small error in representing 0.0001
in binary propagates to the sum.
For the same reason, you should always be very cautious when making
comparisons on real numbers. The following example illustrates a
common programming error:
item1# = 69.82#
item2# = 69.20# + 0.62#
IF item1# = item2# then print "Equality!"
This will NOT PRINT "Equality!" because 69.82 cannot be represented
exactly in binary, which causes the value that results from the
assignment to be SLIGHTLY different (in binary) than the value that is
generated from the expression. In practice, you should always code
such comparisons in such a way as to allow for some tolerance. For
example:
IF (item1# < 69.83#) AND (item1# > 69.81#) then print "Equal"
This will PRINT "Equal".
IEEE Format Numbers
-------------------
QuickBASIC version 3.0 was shipped with an MBF (Microsoft Binary
Floating Point) version and an IEEE (Institute of Electrical and
Electronics Engineers) version for machines with a math coprocessor.
QuickBASIC versions 4.0 and later use IEEE only. Microsoft chose the
IEEE standard to represent floating-point values in current versions
of BASIC for three primary reasons:
1. To allow BASIC to use the Intel math coprocessors, which use IEEE
format. The Intel 80x87 series coprocessors cannot work with
Microsoft Binary Format numbers.
2. To make interlanguage calling between BASIC, C, Pascal, FORTRAN,
and MASM much easier. Otherwise, conversion routines would have to
be used to send numeric values from one language to another.
3. To achieve consistency. IEEE is the accepted industry standard for
C and FORTRAN compilers.
The following is a quick comparison of IEEE and MBF representations
for a double-precision number:
Sign Bits Exponent Bits Mantissa Bits
--------- ------------- -------------
IEEE 1 11 52 + 1 (Implied)
MBF 1 8 56
Note that IEEE has more bits dedicated to the exponent, which allows
it to represent a wider range of values. MBF has more mantissa bits,
which allows it to be more precise within its narrower range.
General Floating-Point Concepts
-------------------------------
It is very important to realize that any binary floating-point system
can represent only a finite number of floating-point values in exact
form. All other values must be approximated by the closest
representable value. The IEEE standard specifies the method for
rounding values to the "closest" representable value. QuickBASIC
supports the standard and rounds according to the IEEE rules.
Also, keep in mind that the numbers that can be represented in IEEE
are spread out over a very wide range. You can imagine them on a
number line. There is a high density of representable numbers near 1.0
and -1.0 but fewer and fewer as you go towards 0 or infinity.
The goal of the IEEE standard, which is designed for engineering
calculations, is to maximize accuracy (to get as close as possible to
the actual number). Precision refers to the number of digits that you
can represent. The IEEE standard attempts to balance the number of
bits dedicated to the exponent with the number of bits used for the
fractional part of the number, to keep both accuracy and precision
within acceptable limits.
IEEE Details
------------
Floating-point numbers are represented in the following form, where
[exponent] is the binary exponent:
X = Fraction * 2^(exponent - bias)
[Fraction] is the normalized fractional part of the number, normalized
because the exponent is adjusted so that the leading bit is always a
1. This way, it does not have to be stored, and you get one more bit
of precision. This is why there is an implied bit. You can think of
this like scientific notation, where you manipulate the exponent to
have one digit to the left of the decimal point, except in binary, you
can always manipulate the exponent so that the first bit is a 1, since
there are only 1s and 0s.
[bias] is the bias value used to avoid having to store negative
exponents.
The bias for single-precision numbers is 127 and 1023 (decimal) for
double-precision numbers.
The values equal to all 0's and all 1's (binary) are reserved for
representing special cases. There are other special cases as well,
that indicate various error conditions.
Single-Precision Examples
-------------------------
2 = 1 * 2^1 = 0100 0000 0000 0000 ... 0000 0000 = 4000 0000h
Note the sign bit is zero, and the stored exponent is 128, or
100 0000 0 in binary, which is 127 plus 1. The stored mantissa is
(1.) 000 0000 ... 0000 0000, which has an implied leading 1 and
binary point, so the actual mantissa is 1.
-2 = -1 * 2^1 = 1100 0000 0000 0000 ... 0000 0000 = C000 0000h
Same as +2 except that the sign bit is set. This is true for all
IEEE format floating-point numbers.
4 = 1 * 2^2 = 0100 0000 1000 0000 ... 0000 0000 = 4080 0000h
Same mantissa, exponent increases by one (biased value is 129, or
100 0000 1 in binary.
6 = 1.5 * 2^2 = 0100 0000 1100 0000 ... 0000 0000 = 40C0 0000 h
Same exponent, mantissa is larger by half -- it's
(1.) 100 0000 ... 0000 0000, which, since this is a binary
fraction, is 1-1/2 (the values of the fractional digits are 1/2,
1/4, 1/8, etc.).
1 = 1 * 2^0 = 0011 1111 1000 0000 ... 0000 0000 = 3F80 0000h
Same exponent as other powers of 2, mantissa is one less than
2 at 127, or 011 1111 1 in binary.
.75 = 1.5 * 2^-1 = 0011 1111 0100 0000 ... 0000 0000 = 3F40 0000h
The biased exponent is 126, 011 1111 0 in binary, and the mantissa
is (1.) 100 0000 ... 0000 0000, which is 1-1/2.
2.5 = 1.25 * 2^1 = 0100 0000 0010 0000 ... 0000 0000 = 4020 0000h
Exactly the same as 2 except that the bit which represents 1/4 is
set in the mantissa.
0.1 = 1.6 * 2^-4 = 0011 1101 1100 1100 ... 1100 1101 = 3DCC CCCDh
1/10 is a repeating fraction in binary. The mantissa is just shy
of 1.6, and the biased exponent says that 1.6 is to be divided by
16 (it is 011 1101 1 in binary, which is 123 in decimal). The true
exponent is 123 - 127 = -4, which means that the factor by which
to multiply is 2**-4 = 1/16. Note that the stored mantissa is
rounded up in the last bit. This is an attempt to represent the
unrepresentable number as accurately as possible. (The reason that
1/10 and 1/100 are not exactly representable in binary is similar
to the way that 1/3 is not exactly representable in decimal.)
0 = 1.0 * 2^-128 = all zeros -- a special case.
Other Common Floating-Point Errors
----------------------------------
The following are common floating-point errors:
1. Round-off error
This error results when all of the bits in a binary number cannot
be used in a calculation.
Example: Adding 0.0001 to 0.9900 (Single Precision)
Decimal 0.0001 will be represented as:
(1.)10100011011011100010111 * 2^(-14+Bias) (13 Leading 0s in
Binary!)
0.9900 will be:
(1.)11111010111000010100011 * 2^(-1+Bias)
Now to actually add these numbers, the decimal (binary) points must
be aligned. For this they must be Unnormalized. Listed below is the
resulting addition:
.000000000000011010001101 * 2^0 <- Only 11 of 23 Bits retained
+.111111010111000010100011 * 2^0
________________________________
.111111010111011100110000 * 2^0
This is called a round-off error because some computers round when
shifting for addition. Others simply truncate. Round-off errors are
important to consider whenever you are adding or multiplying two
very different values.
2. Subtracting two almost equal values
.1235
-.1234
_____
.0001
This will be normalized. Note that although the original numbers
each had four significant digits, the result has only one
significant digit.
3. Overflow and underflow
This occurs when the result is too large or too small to be
represented by the data type.
4. Quantizing error
This occurs with those numbers that cannot be represented in exact
form by the floating-point standard.
5. Division by a very small number
This can trigger a "divide by zero" error or can produce bad
results, as in the following example:
A = 112000000
B = 100000
C = 0.0009
X = A - B / C
In QuickBASIC, X now has the value 888887, instead of the correct
answer, 900000.
6. Output error
This type of error occurs when the output functions alter the
values they are working with.