Bash, non-integers and arithmetic
 

Why is this happening?

$ let total+="4.5"; echo $total
-sh: let: total+=4.5: syntax error in expression (error token is ".5")
4

I need the variable to hold a non-integer. What's the workaround? I've read quite a few tutorials on the net concerning arithmetic and variables and the answer I keep getting is this one:

... yet if I do:

$ let total+=4.5|bc -l; echo $total

I get:

-sh: let: total+=4.5: syntax error in expression (error token is ".5")
4

What is the proper syntax?

Are there more comprehensive tutorials than the ones I have encountered?

The problem is the definition of total. The code will need to echo 3/4 then pipe through bc -l, as:
which returns .75000000000000000000 on my system. (=

Thanks Linux Duck,

How would I get BASH to return only the first two digits after the decimal, essentially giving a result of .75? Is there a command which could restrict output to a "dollar" notation?

i.e. total=.75 or total=2.50 or total=21.25

Thanks again for your help.

Sure! There are several methods of doing so. One is to adjust the scale (number of decimal places that bc keeps). You could change the bc line to:
which returns
That will solve it. Or, printf will do it, and make it a bit more friendly, as far as display money value goes. Change the echo to something like:

which outputs
Just in case you're not familiar with the format, %g is the format for displaying a double, .2 will cause it to retain only the first two decimal places, and the 0 will cause it to be padded with zero's.

That should do it!!

Hello Linux Duck:
I tried
:printf "total is \$%0.2f\n" $total:
The output is
total is $0.00
.75000000000000000000$
Could you please explain this?
Thanking you,
Mons...

mons:

try changing the 'f' to 'g'. Also, can you post the script, just so I can see what it's doing?

TLD