Can a shell be an "interpreter"?

From: trifon (ivaylo_iliev_at_hotmail.com)
Date: 11/17/03 

Date: 16 Nov 2003 21:59:03 -0800

Sorry if this is trivial, couldn't find anything related.

If I have a script named bctest1:

#! /usr/bin/bc -q
2+2
quit
### bctest1 end ###

and run it, the output is:
$ ./bctest1
4

Instead of
#! /usr/bin/bc
I can have
#! /bin/bash,
#! /bin/octave, etc. whatever "interpreter" specified on the first
line. Is it possible the so called interpreter to be another script?
When I run bctest2 which is:

#! ./shellbc -q
2+2
quit
### bctest2 end ###

where shellbc is just:

### shellbc begin ###
/usr/bin/bc $*
### shellbc end ###

I get:
$ ./bctest2
./bctest2: line 2: 2+2: command not found
./bctest2: line 3: quit: command not found

although otherwise ./shellbc will do everything which /usr/bin/bc
would. There is no problem with the permissions, spelling, etc. I get
the same from all "interpreters" I've tryed, wrap them in another
script and can't execute your file.

Thanks for your help...

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