bash expr：非整数参数。如何减去点十进制数
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expr: non-integer argument. How to subtract dot-decimal numbers
提问by Coenster
I am trying to subtract two numbers by running the following bash script:
我试图通过运行以下 bash 脚本来减去两个数字：
#!/bin/bash -x
cur_length=`cat length.txt`
cur_pos=`cat pos.txt`
diff=`$(expr $cur_length - $cur_pos)`
echo "$diff"
But the output says expr has some issue:
但是输出说 expr 有一些问题：
+++ expr 235.68 - 145.9
expr: non-integer argument
+ diff=
+ echo ''
I have searched for "expr: non-integer argument" on the net, but nothing involves dot-decimal numbers. How can I subtract numbers like this? 235.68 - 145.9
我在网上搜索过“expr：非整数参数”，但没有涉及点十进制数。我怎样才能减去这样的数字？235.68 - 145.9
Thanks in advance.
提前致谢。
回答by John1024
Bash doesn't do fractions, just integers. Use bcinstead:
Bash 不做分数，只做整数。使用bc来代替：
$ echo '235.68 - 145.9' | bc
89.78
That result can, of course, be put in a shell variable the same way that you were doing with expr:
当然，该结果可以以与您相同的方式放入 shell 变量中expr：
$ diff="$(echo '235.68 - 145.9' | bc)"
$ echo $diff
89.78
