浮点导致Bash整数除法 [英] Floating point results in Bash integer division

问题描述

我的服务器上有一个备份脚本，该脚本执行备份作业，并向我发送一份备份文件的摘要，包括新备份文件的大小.作为脚本的一部分，我想将文件的最终大小除以(1024 ^ 3)，以从文件大小(以字节为单位)获得以GB为单位的文件大小.

I have a backup script on my server which does cron jobs of backups, and sends me a summary of files backed up, including the size of the new backup file. As part of the script, I'd like to divide the final size of the file by (1024^3) to get the file size in GB, from the file size in bytes.

由于bash没有浮点计算，因此我尝试使用管道对bc进行计算以获得结果，但是我对基本示例感到困惑.

Since bash does not have floating point calculation, I am trying to use pipes to bc to get the result, however I'm getting stumped on basic examples.

但是，我试图将Pi的值调整为一个比例，

I tried to get the value of Pi to a scale, however,

即使以下方法可行:

~ #bc -l
bc 1.06.95
Copyright 1991-1994, 1997, 1998, 2000, 2004, 2006 Free Software Foundation, Inc.
This is free software with ABSOLUTELY NO WARRANTY.
For details type `warranty'.
4/3
1.33333333333333333333
22/7
3.14285714285714285714
q
0
quit

非交互式版本不起作用:

A non interactive version does not work:

#echo $(( 22/7 )) | bc
3

这有效:

#echo '22/7' | bc -l
3.14285714285714285714

但是我需要使用变量.因此，以下操作无效无济于事:

But I need to use variables. So it doesnt help that the following does not work:

#a=22 ; b=7
#echo $(( a/b )) | bc -l
3

我显然在Bash中使用变量的语法中缺少某些内容，并且可能与一些指针"一起使用来误解我的内容.

I'm obviously missing something in the syntax for using variables in Bash, and could use with some 'pointers' on what I've misunderstood.

正如DigitalRoss所说，我可以使用以下内容:

As DigitalRoss said, I can use the following:

#echo $a / $b | bc -l
3.14285714285714285714

但是我不能使用复杂的表达式，例如:

However I cant use complex expressions like:

#echo $a / (( $b-34 )) | bc -l
-bash: syntax error near unexpected token `('
#echo $a / (( b-34 )) | bc -l
-bash: syntax error near unexpected token `('
#echo $a / (( b-34 )) | bc -l
-bash: syntax error near unexpected token `('

有人可以给我一种正确的语法，以便通过复杂的算术表达式获得浮点结果吗?

Can someone give me a working correct syntax for getting floating point results with complicated arithmetic expresssions?

推荐答案

只需对表达式进行双引号(")

Just double-quote (") the expression:

echo "$a / ( $b - 34 )" | bc -l

然后bash将扩展$变量并忽略其他所有内容，并且bc将看到带括号的表达式:

Then bash will expand the $ variables and ignore everything else and bc will see an expression with parentheses:

$ a=22
$ b=7
$ echo "$a / ( $b - 34 )" 
22 / ( 7 - 34 )

$ echo "$a / ( $b - 34 )" | bc -l
-.81481481481481481481

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