本文介绍了为什么我在Perl中减去同一个浮点数的时候得不到零呢？的处理方法，对大家解决问题具有一定的参考价值，需要的朋友们下面随着小编来一起学习吧！
问题描述
＃！/ usr / bin / perl
$ l1 =0 + 0.590580 + 0.583742 + 0.579787 + 0.564928 + 0.504538 + 0.459805 + 0.433273 + 0.384211 + 0.3035810;
$ l2 =0 + 0.590580 + 0.583742 + 0.579788 + 0.564928 + 0.504538 + 0.459805 + 0.433272 + 0.384211 + 0.3035810;
$ val1 = eval（$ l1）;
$ val2 = eval（$ l2）;
$ diff =（（$ val1 - $ val2）/ $ val1）* 100; $（$ val1 - $ val2）/ $ val1）* 100 ==> $ diff \ n;
令人惊讶的是，输出结果是
<$ （（4.404445 - 4.404445）/4.404445）* 100 ==> -2.01655014354845e-14。
是不是应该是ZERO？
任何人都可以解释这个请求......
解决方案
这不是Perl相关的，而是浮点相关的。
#!/usr/bin/perl
$l1 = "0+0.590580+0.583742+0.579787+0.564928+0.504538+0.459805+0.433273+0.384211+0.3035810";
$l2 = "0+0.590580+0.583742+0.579788+0.564928+0.504538+0.459805+0.433272+0.384211+0.3035810";
$val1 = eval ($l1);
$val2 = eval ($l2);
$diff = (($val1 - $val2)/$val1)*100;
print " (($val1 - $val2)/$val1)*100 ==> $diff\n";
Surprisingly the output ended up to be
((4.404445 - 4.404445)/4.404445)*100 ==> -2.01655014354845e-14.
Is it not supposed to be a ZERO????Can any one explain this please......
解决方案
What every computer scientist should know about floating point arithmetic
See Why is floating point arithmetic in C# imprecise?
This isn't Perl related, but floating point related.
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