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PERLFAQ4(1) Perl Programmers Reference Guide PERLFAQ4(1)NAMEperlfaq4 - Data Manipulation ($Revision: 10394 $)DESCRIPTIONThis section of the FAQ answers questions related to manipulating numbers, dates, strings, arrays, hashes, and miscellaneous data issues. Data: Numbers Why am I getting long decimals (eg, 19.9499999999999) instead of the numbers I should be getting (eg, 19.95)? Internally, your computer represents floating-point numbers in binary. Digital (as in powers of two) computers cannot store all numbers exactly. Some real numbers lose precision in the process. This is a problem with how computers store numbers and affects all computer languages, not just Perl. perlnumber shows the gory details of number representations and conversions. To limit the number of decimal places in your numbers, you can use the printf or sprintf function. See the "Floating Point Arithmetic" for more details. printf "%.2f", 10/3; my $number = sprintf "%.2f", 10/3; Why is int() broken? Your "int()" is most probably working just fine. It's the numbers that aren't quite what you think. First, see the answer to "Why am I getting long decimals (eg, 19.9499999999999) instead of the numbers I should be getting (eg, 19.95)?". For example, this print int(0.6/0.2-2), " "; will in most computers print 0, not 1, because even such simple numbers as 0.6 and 0.2 cannot be presented exactly by floating-point num- bers. What you think in the above as 'three' is really more like 2.9999999999999995559. Why isn't my octal data interpreted correctly? Perl only understands octal and hex numbers as such when they occur as literals in your program. Octal literals in perl must start with a leading 0 and hexadecimal literals must start with a leading "0x". If they are read in from somewhere and assigned, no automatic conver- sion takes place. You must explicitly use "oct()" or "hex()" if you want the values converted to decimal. "oct()" interprets hexadecimal(0x350), octal (0350 or even without the leading 0, like 377) and binary ("0b1010") numbers, while "hex()" only converts hexadecimal ones, with or without a leading "0x", such as 0x255, "3A", "ff", or "deadbeef". The inverse mapping from decimal to octal can be done with either the <%o> or %O "sprintf()" formats. This problem shows up most often when people try using "chmod()", "mkdir()", "umask()", or "sysopen()", which by widespread tradition typi- cally take permissions in octal. chmod(644, $file); # WRONG chmod(0644, $file); # right Note the mistake in the first line was specifying the decimal literal 644, rather than the intended octal literal 0644. The problem can be seen with: printf("%#o",644); # prints 01204 Surely you had not intended "chmod(01204, $file);" - did you? If you want to use numeric literals as arguments to chmod() et al. then please try to express them as octal constants, that is with a leading zero and with the following digits restricted to the set 0..7. Does Perl have a round() function? What about ceil() and floor()? Trig functions? Remember that "int()" merely truncates toward 0. For rounding to a certain number of digits, "sprintf()" or "printf()" is usually the eas- iest route. printf("%.3f", 3.1415926535); # prints 3.142 The "POSIX" module (part of the standard Perl distribution) implements "ceil()", "floor()", and a number of other mathematical and trigono- metric functions. use POSIX; $ceil = ceil(3.5); # 4 $floor = floor(3.5); # 3 In 5.000 to 5.003 perls, trigonometry was done in the "Math::Complex" module. With 5.004, the "Math::Trig" module (part of the standard Perl distribution) implements the trigonometric functions. Internally it uses the "Math::Complex" module and some functions can break out from the real axis into the complex plane, for example the inverse sine of 2. Rounding in financial applications can have serious implications, and the rounding method used should be specified precisely. In these cases, it probably pays not to trust whichever system rounding is being used by Perl, but to instead implement the rounding function you need yourself. To see why, notice how you'll still have an issue on half-way-point alternation: for ($i = 0; $i < 1.01; $i += 0.05) { printf "%.1f ",$i} 0.0 0.1 0.1 0.2 0.2 0.2 0.3 0.3 0.4 0.4 0.5 0.5 0.6 0.7 0.7 0.8 0.8 0.9 0.9 1.0 1.0 Don't blame Perl. It's the same as in C. IEEE says we have to do this. Perl numbers whose absolute values are integers under 2**31 (on 32 bit machines) will work pretty much like mathematical integers. Other numbers are not guaranteed. How do I convert between numeric representations/bases/radixes? As always with Perl there is more than one way to do it. Below are a few examples of approaches to making common conversions between num- ber representations. This is intended to be representational rather than exhaustive. Some of the examples later in perlfaq4 use the "Bit::Vector" module from CPAN. The reason you might choose "Bit::Vector" over the perl built in functions is that it works with numbers of ANY size, that it is optimized for speed on some operations, and for at least some pro- grammers the notation might be familiar. How do I convert hexadecimal into decimal Using perl's built in conversion of "0x" notation: $dec = 0xDEADBEEF; Using the "hex" function: $dec = hex("DEADBEEF"); Using "pack": $dec = unpack("N", pack("H8", substr("0" x 8 . "DEADBEEF", -8))); Using the CPAN module "Bit::Vector": use Bit::Vector; $vec = Bit::Vector->new_Hex(32, "DEADBEEF"); $dec = $vec->to_Dec(); How do I convert from decimal to hexadecimal Using "sprintf": $hex = sprintf("%X", 3735928559); # upper case A-F $hex = sprintf("%x", 3735928559); # lower case a-f Using "unpack": $hex = unpack("H*", pack("N", 3735928559)); Using "Bit::Vector": use Bit::Vector; $vec = Bit::Vector->new_Dec(32, -559038737); $hex = $vec->to_Hex(); And "Bit::Vector" supports odd bit counts: use Bit::Vector; $vec = Bit::Vector->new_Dec(33, 3735928559); $vec->Resize(32); # suppress leading 0 if unwanted $hex = $vec->to_Hex(); How do I convert from octal to decimal Using Perl's built in conversion of numbers with leading zeros: $dec = 033653337357; # note the leading 0! Using the "oct" function: $dec = oct("33653337357"); Using "Bit::Vector": use Bit::Vector; $vec = Bit::Vector->new(32); $vec->Chunk_List_Store(3, split(//, reverse "33653337357")); $dec = $vec->to_Dec(); How do I convert from decimal to octal Using "sprintf": $oct = sprintf("%o", 3735928559); Using "Bit::Vector": use Bit::Vector; $vec = Bit::Vector->new_Dec(32, -559038737); $oct = reverse join('', $vec->Chunk_List_Read(3)); How do I convert from binary to decimal Perl 5.6 lets you write binary numbers directly with the "0b" notation: $number = 0b10110110; Using "oct": my $input = "10110110"; $decimal = oct( "0b$input" ); Using "pack" and "ord": $decimal = ord(pack('B8', '10110110')); Using "pack" and "unpack" for larger strings: $int = unpack("N", pack("B32", substr("0" x 32 . "11110101011011011111011101111", -32))); $dec = sprintf("%d", $int); # substr() is used to left pad a 32 character string with zeros. Using "Bit::Vector": $vec = Bit::Vector->new_Bin(32, "11011110101011011011111011101111"); $dec = $vec->to_Dec(); How do I convert from decimal to binary Using "sprintf" (perl 5.6+): $bin = sprintf("%b", 3735928559); Using "unpack": $bin = unpack("B*", pack("N", 3735928559)); Using "Bit::Vector": use Bit::Vector; $vec = Bit::Vector->new_Dec(32, -559038737); $bin = $vec->to_Bin(); The remaining transformations (e.g. hex -> oct, bin -> hex, etc.) are left as an exercise to the inclined reader. Why doesn't & work the way I want it to? The behavior of binary arithmetic operators depends on whether they're used on numbers or strings. The operators treat a string as a series of bits and work with that (the string "3" is the bit pattern 00110011). The operators work with the binary form of a number (the number 3 is treated as the bit pattern 00000011). So, saying "11 & 3" performs the "and" operation on numbers (yielding 3). Saying "11" & "3" performs the "and" operation on strings (yielding "1"). Most problems with "&" and "|" arise because the programmer thinks they have a number but really it's a string. The rest arise because the programmer says: if ("