「無效的算術運算符」在bash做浮點運算

Question

這裏是我的腳本，其相當不言自明：

d1=0.003 
d2=0.0008 
d1d2=$((d1 + d2)) 

mean1=7 
mean2=5 
meandiff=$((mean1 - mean2)) 

echo $meandiff 
echo $d1d2 

但是，而不是領我預期的輸出： 0.0038 我得到錯誤Invalid Arithmetic Operator, (error token is ".003")?

Answer 1

bash不支持浮點運算。您需要使用外部工具，如bc。

# Like everything else in shell, these are strings, not 
# floating-point values 
d1=0.003 
d2=0.0008 

# bc parses its input to perform math 
d1d2=$(echo "$d1 + $d2" | bc) 

# These, too, are strings (not integers) 
mean1=7 
mean2=5 

# $((...)) is a built-in construct that can parse 
# its contents as integers; valid identifiers 
# are recursively resolved as variables. 
meandiff=$((mean1 - mean2))