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Home/ Questions/Q 9118823
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Editorial Team
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Editorial Team
Asked: June 17, 20262026-06-17T05:14:48+00:00 2026-06-17T05:14:48+00:00

How to do arithmetic with floating point numbers such as 1.503923 in a shell

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How to do arithmetic with floating point numbers such as 1.503923 in a shell script? The floating point numbers are pulled from a file as a string. The format of the file is as follows:

1.5493482,3.49384,33.284732,23.043852,2.2384...
3.384,3.282342,23.043852,2.23284,8.39283...
.
.
.

Here is some simplified sample code I need to get working. Everything works fine up to the arithmetic. I pull a line from the file, then pull multiple values from that line. I think this would cut down on search processing time as these files are huge.

# set vars, loops etc.

while [ $line_no -gt 0 ]
do
    line_string=`sed -n $line_no'p' $file_path`  # Pull Line (str) from a file
    string1=${line_string:9:6}                   # Pull value from the Line
    string2=${line_string:16:6}
    string3=...
    .
    .
    .
    calc1= `expr $string2 - $string7` |bc -l     # I tried these and various
    calc2= ` "$string3" * "$string2" ` |bc -l    # other combinations
    calc3= `expr $string2 - $string1`
    calc4= "$string2 + $string8" |bc
    .
    .
    .
    generic_function_call                        # Use the variables in functions
    line_no=`expr $line_no - 1`                  # Counter--
done

Output I keep getting:

expr: non-numeric argument
command not found
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1 Answer

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  1. Editorial Team
    Editorial Team
    2026-06-17T05:14:49+00:00Added an answer on June 17, 2026 at 5:14 am

    What about this?

    calc=$(echo "$String2 + $String8"|bc)
    

    This will make bc to add the values of $String2 and $String8 and saves the result in the variable calc.

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