在perl中舍入十进制数，错误的结果 [英] Rounding decimal numbers in perl, wrong result

问题描述

我讨厌十进制数字.对于 1.005，我没有得到以下代码所期望的结果.

I hate decimal numbers. For 1.005 I don't get the result I expect with the following code.

#!/usr/bin/perl -w

use strict;
use POSIX qw(floor);

my $num = (1.005 * 100) + 0.5;
print $num . "
";          # 101
print floor($num) . "
";   # 100
print int($num) . "
";     # 100

对于 2.005 和 3.005，它工作正常.

For 2.005 and 3.005 it works fine.

通过这个丑陋的hack"，我得到了预期的结果.

With this ugly "hack" I get the expected result.

#!/usr/bin/perl -w

use strict;
use POSIX qw(floor);

my $num = (1.005 * 100) + 0.5;
$num = "$num";
print $num . "
";          # 101
print floor($num) . "
";   # 101
print int($num) . "
";     # 101

这样做的正确方法是什么?

What is the correct way to do this?

推荐答案

floor() 不是四舍五入，而是向下取最接近的整数.

floor() is not for rounding, it goes down to the nearest integer.

查看这篇旧帖子:你如何舍入Perl 中的浮点数?

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