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in reply to Simple math gone wrong

$ perldoc -q decimal Found in /opt/perl/lib/5.8.0/pod/perlfaq4.pod Why am I getting long decimals (eg, 19.9499999999999) instead of the numbers I should be getting (eg, 19.95)? The infinite set that a mathematician thinks of as the real numbers can only be approximated on a computer, since the computer only has a finite number of bits to store an infinite number of, um, numbers. Internally, your computer represents floating- point numbers in binary. Floating-point numbers read in from a file or appearing as literals in your program are converted from their decimal floating-point representation (eg, 19.95) to an internal binary representation. However, 19.95 can't be precisely represented as a binary floating-point number, just like 1/3 can't be exactly represented as a decimal floating-point number. The computer's binary representation of 19.95, therefore, isn't exactly 19.95. When a floating-point number gets printed, the binary floating-point representation is converted back to decimal. These decimal numbers are dis- played in either the format you specify with printf(), or the current output format for num- bers. (See "$#" in perlvar if you use print. $# has a different default value in Perl5 than it did in Perl4. Changing $# yourself is deprecated.) This affects all computer languages that represent decimal floating-point numbers in binary, not just Perl. Perl provides arbitrary-precision decimal numbers with the Math::BigFloat module (part of the standard Perl distribution), but mathematical operations are consequently slower. If precision is important, such as when dealing with money, it's good to work with integers and then divide at the last possible moment. For example, work in pennies (1995) instead of dollars and cents (19.95) and divide by 100 at the end. To get rid of the superfluous digits, just use a format (eg, "printf("%.2f", 19.95)") to get the required precision. See "Floating-point Arith- metic" in perlop.

Abigail