On 2015-10-29, staticx <mohamedboussaa3@xxxxxxxxx> wrote: > I wonder if I can pass the content of a bash variable as an argument to gcc. > My variable is a list of optimization options like $var="-opt1 -opt2 -opt3" > gcc $var test.c -o object.o didn't work What do you mean by `didn't work'? This worked for me: $ ll total 4.0K -rw-r--r-- 1 ja ja 405 Oct 29 14:33 main.c $ export VAR="main.c -o PROG -O2 -g" $ gcc $VAR $ ll total 16K -rw-r--r-- 1 ja ja 405 Oct 29 14:33 main.c -rwxr-xr-x 1 ja ja 11K Oct 29 14:43 PROG You just can't do: $ VAR="main.c -o PROG -O2 -g" gcc $VAR gcc: fatal error: no input files compilation terminated. because $VAR will expanded at the very beginning before it is defined and will be empty. This not gcc problem, it's how Bash works. -- Arkadiusz Drabczyk <arkadiusz@xxxxxxxxxxxx>
