[Haskell-cafe] GHC RTS question
Daniel Fischer
daniel.is.fischer at web.de
Sun Feb 21 14:12:11 EST 2010
Am Sonntag 21 Februar 2010 19:58:12 schrieb Artyom Kazak:
> 2010/2/21 Daniel Fischer <daniel.is.fischer at web.de>:
> > Am Sonntag 21 Februar 2010 18:20:43 schrieb Artyom Kazak:
> >> > Enclose it in double quotes (perhaps single quotes would also work)
> >>
> >> No, I want my program to work the same way as UNIX "echo" does.
> >> Without any double quotes.
> >
> > Okay, what about
> >
> > "If you absolutely positively want all the rest of the options in a
> > command line to go to the program (and not the RTS), use a ––RTS."
> >
> > $ ./prog +RTS --RTS +RTS
> >
> > ? (BTW, enclosing in quotes doesn't work anyway if the argument
> > consists *only* of "+RTS", same as with echo, echo "-e" doesn't output
> > '-e' either).
>
> So, if I type "./prog +RTS --RTS +RTS", the output will be "+RTS". But
> I want the output to be equal to the input IN ALL CASES, without any
> quotes, additional options, etc. I want all the command line to go to
> my program. How can I do it? (The only way I know now - hacking the
> GHC. If there are no other ways, I'll do it.)
Shell wrapper:
$ cat wopTest
./opTest +RTS --RTS $@
$ cat opTest.hs
module Main (main) where
import System.Environment (getArgs)
main = mapM_ print =<< getArgs
$ ./wopTest +RTS -sstderr -RTS
"+RTS"
"-sstderr"
"-RTS"
Other than that, hacking GHC is the only way I can think of either, since
looking for RTS-options is a fixed (and generally necessary) part of the
RTS.
But why do you want that behaviour so much that you'd be willing to hack
GHC?
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