Passing parameter through Shell to C

Question

name="Rahul Kapgate arg"
temp=$name
./a.out -a "$temp"

This will give me a following output :: Rahul Kapgate arg

Instead of ./a.out -a "$temp" if i put following logic then it is generating differant output.

cmd="./a.out -a \"$temp\""
$cmd

Output :: "Rahul

[ a.out is nothing but the printing the 1st parameter through C unsing optarg())

I am expecting to generate the same output using 2nd option plz help me.

Answer 1

Bash FAQ entry #50

Answer 2

It seems you are bypassing some of the shell processing. Try this:

echo $cmd | bash -s

Answer 3

As the information in the link provided by Ignacio indicates, you should try to avoid putting commands in variables. However, this may work for you (based also on information there):

cmd="./a.out"
args=(-a "$temp")
$cmd "${args[@]}"