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Re: Cannot pass a single backslash in multiboot cmdline

From: Jakub Jermář
Subject: Re: Cannot pass a single backslash in multiboot cmdline
Date: Mon, 26 Dec 2016 19:56:44 +0100
User-agent: Mozilla/5.0 (X11; Linux x86_64; rv:45.0) Gecko/20100101 Thunderbird/45.4.0

On 12/26/2016 07:24 PM, Andrei Borzenkov wrote:
> 26.12.2016 21:12, Jakub Jermář пишет:
>>>> I am observing a strange behavior when passing boot arguments with a
>>>> backslash to the kernel (the multiboot cmd_line via the multiboot
>>>> command in grub.cfg). I would like to pass foo\bar to the kernel, but to
>>>> no avail. I tried:
>>> Which kernel? What do you load?
>> The kernel is a modified version of HelenOS, but IMHO this issue is
>> kernel agnostic.
> Not really. It depends on target program being loaded which is why I
> asked what you use.

Ok, I submitted my grub.cfg and stated that HelenOS is
multiboot-compliant, gets loaded by the multiboot command. We boot from
an ISO image, which was created like this:

"The binary files of GRUB boot loader in this directory have been
created by compiling GRUB for the 'i386-pc' target and then using the
grub-mkrescue script to create the El Torito boot image."

Is there anything else I should include in order to answer your question?

>> The cmd_line is already wrong when picked up from the
>> multiboot info and printed out (so there is no processing on it from the
>> HelenOS side).
> Sure, but /if/ your kernel interpreted `\' as quoting character, the
> result would be correct.

But instead of the plain actual data there would be escaped data, which
is a hassle. And also a possible workaround.

>> I think I understand why it is eating the single backslash (expected
>> behavior),
> You have two levels here. First level is grub command line processing.
> It is loosely compatible with (Bourne) shell so yes, backslash is
> treated as escape character. You can still pass backslash using usual
> shell quoting, e.g. foo\\bar or 'foo\bar'. Second level is
> grub_create_loader_cmdline() function. This one escapes `\' with second
> `\'. So in the above case this function receives `foo\bar' as adds
> second `\'.

This explains a lot, but does not make much sense for my usecase.

Thank you,

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