当我对引用变量和非引用变量进行数学运算时,为什么会得到不同的结果？

Question

我正在处理纬度和经度来确定商业地点并遇到一些奇怪的行为.

在下面的Perl片段中,等式分配数据以$v1评估为1.当我调用时acos($v1),我收到sqrt错误.当我打电话acos("$v1")(带引号)时,我没有.调用acos(1)也不会产生错误.为什么报价很重要？

use strict;
use warnings 'all';

sub acos {
    my $rad = shift;
    return (atan2(sqrt(1 - $rad**2), $rad));
}

my $v1 = (0.520371764072297 * 0.520371764072297) +
         (0.853939826425894 * 0.853939826425894 * 1);

print acos($v1);   # Can't take sqrt of -8.88178e-16 at foo line 8.
print acos("$v1"); # 0

Answer 1

$v1 不完全是1:

$ perl -e'
    $v1 = (0.520371764072297 * 0.520371764072297) +
          (0.853939826425894 * 0.853939826425894 * 1);
    printf "%.16f\n", $v1
'
1.0000000000000004

但是,当你对它进行字符串化时,Perl只保留15位数的精度:

$ perl -MDevel::Peek -e'
    $v1 = (0.520371764072297 * 0.520371764072297) +
          (0.853939826425894 * 0.853939826425894 * 1);
    Dump "$v1"
'
SV = PV(0x2345090) at 0x235a738
  REFCNT = 1
  FLAGS = (PADTMP,POK,pPOK)
  PV = 0x2353980 "1"\0      # string value is exactly 1
  CUR = 1
  LEN = 16