In bash,I am trying to do math with a integer and a float number to get a integer result. below code snippet doesn't work:
x=25
y=0.2
z=$((x*y))
echo $x*$y=$z
The error message is:
sh: line 3: 0.2: syntax error: invalid arithmetic operator (error token is ".2")
If both variable is integer, it works fine.
How can I get "25*0.2=5" from bash script?
Place your printout in quotes in echo. Also your z=$((x*y)) will make z empty or error:
25*0.2: syntax error: invalid arithmetic operator (error token is ".2")
So... Here is tested code and might be like this:
x=25
y=0.2
z=$(echo $x*$y | bc)
echo "$x*$y=$z"
result will be like this:
25*0.2=5.0
Note: we used bc command for z calculation
Try bashj (a bash mutant with java support) https://sourceforge.net/projects/bashj/.
for instance:
#!/usr/bin/bashj
echo Math.cos(0.5)
echo Math.hypot(3.0,4.0)
bashdoesn't support floating point arithmetic. Usebcorawkorperl