在bash中修复整数长度时的奇怪行为 [英] strange behavior when fixing integer lengths in bash
问题描述
在我的bash脚本中,我设置了以下操作
In my bash script, I set-up the following operations
year=0050
echo $(printf %04d $year)
>0040
我不明白为什么会返回0040 0050.我最终发现要让系统正确打印0050,我将不得不这样做。
I do not understand why 0040 is returned instead of 0050. I eventually found that to get the system to print 0050 correctly, I will have to do this instead.
year=50
echo $(printf %04d $year)
>0050
有没有关于为什么第一种情况发生的见解?
Are there any insights as to why the first case happens?
推荐答案
这是因为带有前导零的数字被Bash解释为八进制,八进制50是十进制40。
It's because numbers with a leading zero are interpreted as octal by Bash, and octal 50 is decimal 40.
要修复它,你可以使用参数扩展来剥离零:
To fix it, you can either strip the zeros with a parameter expansion:
$ printf '%04d\n' "${year##+(0)}"
0050
我放弃了 echo $(...)
构造并在格式中插入换行符而不是字符串。
I've dropped the echo $(...)
construct and inserted a newline in the formatting string instead.
请注意 +(0)
模式需要 \\ textglob
shell选项( shopt -s \\ textglob
)。
Notice that the +(0)
pattern requires the extglob
shell option (shopt -s extglob
).
或者(更便携),您可以先用算术扩展转换数字:
Alternatively (and more portably), you can convert the number with an arithmetic expansion first:
% printf '%04d\n' "$(( 10#$year ))"
0050
这使用 base # n
表示 n
(在我们的例子中: $ year
)的表示法是基数10而不是八进制。
This uses the base#n
notation to indicate that n
(in our case: $year
) is in base 10 and not octal.
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