Hashbang won't support parameters with arguments

Question

Possible Duplicate:
how to use multiple arguments with a shebang (i.e. #!)?

How can I make a #! statement accept a param with arguments? It seems to be lumping them all together as one param instead of splitting on space as usual.

Take this contrived example:

$ cat /tmp/echo
#!/bin/echo -n
$ /tmp/echo
/tmp/echo$

works great and outputs the filename without a new line at the end. But this one:

$ cat /tmp/echo
#!/bin/echo -n hi
$ /tmp/echo
-n hi /tmp/echo
$

kills the -n arg.

One hack I can do is make another shell script that just execs the first script I want with the params I want, but I'd rather not add extra dependancies if I don't need them.

For me, the first example works like yours and the second works like you intended (Mac OS X Mountain Lion, bash 3.2.48(1)) — Pascal Cuoq, Dec 08 '12 at 20:38

score 2 · Answer 1 · edited May 23 '17 at 12:12

2

There apparently isn't a POSIX standard for this, but many (possibly most) *nix systems only allow you to send one argument to the program invoked in the shabang. So using #!/bin/echo -n hi is effectively the same as running /bin/echo "-n hi" /path/to/script.

See this SO answer for a more thorough explanation.

edited May 23 '17 at 12:12

Community

1
1

answered Dec 08 '12 at 20:46

Tim Pote

27,191
6
63
65

Hashbang won't support parameters with arguments

1 Answers1