Next: Try To Round, Previous: Inexactness of computations, Up: FP Math Caution [Contents][Index]
Can arbitrary-precision arithmetic give exact results? There are no easy answers. The standard rules of algebra often do not apply when using floating-point arithmetic. Among other things, the distributive and associative laws do not hold completely, and order of operation may be important for your computation. Rounding error, cumulative precision loss, and underflow are often troublesome.
When gawk
tests the expressions ‘0.1 + 12.2’ and
‘12.3’ for equality using the machine double-precision arithmetic,
it decides that they are not equal! (See section Be Careful Comparing Values.)
You can get the result you want by increasing the precision; 56 bits in
this case does the job:
$ gawk -M -v PREC=56 'BEGIN { print (0.1 + 12.2 == 12.3) }' -| 1
If adding more bits is good, perhaps adding even more bits of
precision is better?
Here is what happens if we use an even larger value of PREC
:
$ gawk -M -v PREC=201 'BEGIN { print (0.1 + 12.2 == 12.3) }' -| 0
This is not a bug in gawk
or in the MPFR library.
It is easy to forget that the finite number of bits used to store the value
is often just an approximation after proper rounding.
The test for equality succeeds if and only if all bits in the two operands
are exactly the same. Because this is not necessarily true after floating-point
computations with a particular precision and effective rounding mode,
a straight test for equality may not work. Instead, compare the
two numbers to see if they are within the desirable delta of each other.
In applications where 15 or fewer decimal places suffice, hardware double-precision arithmetic can be adequate, and is usually much faster. But you need to keep in mind that every floating-point operation can suffer a new rounding error with catastrophic consequences, as illustrated by our earlier attempt to compute the value of pi. Extra precision can greatly enhance the stability and the accuracy of your computation in such cases.
Additionally, you should understand that repeated addition is not necessarily equivalent to multiplication in floating-point arithmetic. In the example in Errors Accumulate:
$ gawk 'BEGIN { > for (d = 1.1; d <= 1.5; d += 0.1) # loop five times (?) > i++ > print i > }' -| 4
you may or may not succeed in getting the correct result by choosing
an arbitrarily large value for PREC
. Reformulation of
the problem at hand is often the correct approach in such situations.
Next: Try To Round, Previous: Inexactness of computations, Up: FP Math Caution [Contents][Index]