perl实现16进制数转换成10进制数

#!/usr/bin/perl use warnings; use strict; use Math::BigInt; my %hex_hash = (0=>0,1=>1,2=>2,3=>3,4=>4,5=>5,6=>6,??7=>7,8=>8,9=>9,a=>10,b=>11,c=>12,d=>13,e=>14,f=>15); ???? sub hex_to_dec{ ??my $data = shift; ??my $index = shift; ??--$index; ??my $result = Math::BigInt->new(0); ??my $factor = Math::BigInt->new(1); ??while($index >=2 ){ ????my $digit = substr($data,$index,1); ????if ($digit =~/[a-fA-F]/){ ??????$digit = lc $digit; ??????$digit = $hex_hash{$digit}; ????} ????my $temp = Math::BigInt->new($digit); ????$temp->bmul($factor); ????$result->badd($temp); ????--$index; ????$factor->bmul(16); ??} ??return $result; } print hex_to_dec("0x10",length("0x10")),"n"; print hex_to_dec("0xfff",length("0xfff")),"n";