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

问题描述

我讨厌十进制数字.对于1.005，使用以下代码无法得到预期的结果.

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 $num . "\n";          # 101
print floor($num) . "\n";   # 100
print int($num) . "\n";     # 100

对于2.005和3.005，它可以正常工作.

For 2.005 and 3.005 it works fine.

有了这个难看的"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 $num . "\n";          # 101
print floor($num) . "\n";   # 101
print int($num) . "\n";     # 101

正确的方法是什么?

推荐答案

floor()不用于四舍五入，它会向下舍入到最接近的整数.

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

请参阅此旧帖子:如何进行舍入Perl中的浮点数?

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