I would like to pass an argument to a function and return a calculated value from it to be stored for further process. Below is the sample code.
#!/bin/bash
test()
{
echo $1
c=$(expr $1 + "10000")
return $c
}
var=$(test 10)
echo $var
I would like to get the value of c stored in var. Can anyone please help in this case.
The "return value" of a function as you used it is stdout.
"return" will set exit status ($?), which you probably have no use for.
"test" is probably a bad choice of name, since it's taken (qv. man test).
So:
$ Test() { expr $1 + 10000; }
$ var=$(Test 10)
$ echo $var
10010
if all you wish to do is add 10000 to your input, then a function is overkill. for this, wouldnt this work?
your_arg=10
var=$(( ${your_arg}+10000 ))
echo $var
There are some issues in your code.
#!/bin/bash
It works but it is no good idea to define a function called test, because test is a standard Unix program.
test()
{
Write debug output to standard error (&2) instead of standard output (&1). Standard output is used to return data.
echo "$1" >&2
Declare variables in functions with local to avoid side effects.
local c=$(expr "$1" + "10000")
Use echo instead of return to return strings. return can return only integers.
echo "$c"
}
var=$(test 10)
If unsure always quote arguments.
echo "$var" >&2
