perl -Mbigrat -E'for (1..100) { $i += 1/3; say int($i), "\t", sprintf "%.55f", $i }'
垃圾邮件很多警告:
Argument "100/3" isn't numeric in addition (+) at …/site_perl/5.24.1/Math/BigRat.pm line 1939.
不带-Mbigrat
再次运行,看看sprintf的预期效果。
如何降级Math::BigRat
将 $i
实例转换为普通 NV令 sprintf
满意吗?
版本:
bigrat
0.47数学::BigRat
0.2612
最佳答案
我期望 Math::BigRat 提供一种获取普通数字的方法,它确实做到了:
as_int
用于整数as_float
用于 float
出于调试目的,我插入了 $_
的打印。
$i->as_int()
与 int($i)
的作用相同:
$ perl -Mbigrat -E'for (1..100) { $i += 1/3; say $_, "\t", int($i), "\t", sprintf "%.55f", $i->as_int() }'
...
99 33 33.0000000000000000000000000000000000000000000000000000000
100 33 33.0000000000000000000000000000000000000000000000000000000
$i->as_float()
乍一看似乎按预期工作,但我不明白输出。所有小数位均为零,并且每隔一行 $i
等于 $_
:
$ perl -Mbigrat -E'for (1..100) { $i += 1/3; say $_, "\t", int($i), "\t", sprintf "%.55f", $i->as_float() }'
...
90 30 30.0000000000000000000000000000000000000000000000000000000
91 30 91.0000000000000000000000000000000000000000000000000000000
92 30 92.0000000000000000000000000000000000000000000000000000000
93 31 31.0000000000000000000000000000000000000000000000000000000
94 31 94.0000000000000000000000000000000000000000000000000000000
95 31 95.0000000000000000000000000000000000000000000000000000000
96 32 32.0000000000000000000000000000000000000000000000000000000
97 32 97.0000000000000000000000000000000000000000000000000000000
98 32 98.0000000000000000000000000000000000000000000000000000000
99 33 33.0000000000000000000000000000000000000000000000000000000
100 33 100.0000000000000000000000000000000000000000000000000000000
这是带有 Math::BigRat 0.2614 的 Perl 5.30.0。
因此警告已修复,但此解决方案似乎有问题。
更新:按照评论中的要求,不带sprintf
:
$ perl -Mbigrat -E'for (1..100) { $i += 1/3; say $_, "\t", int($i), "\t", $i->as_float() }'
...
90 30 30
91 30 91
92 30 92
93 31 31
94 31 94
95 31 95
96 32 32
97 32 97
98 32 98
99 33 33
100 33 100
关于perl - 怎么降级大佬?,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/44211409/