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Re: How to compare $arg0 with string literal?
- From: Doug Evans <dje at google dot com>
- To: Aleksey Midenkov <midenok at gmail dot com>
- Cc: "gdb at sourceware dot org" <gdb at sourceware dot org>
- Date: Thu, 26 Mar 2015 11:28:13 -0700
- Subject: Re: How to compare $arg0 with string literal?
- Authentication-results: sourceware.org; auth=none
- References: <CAF8BazAOQC-otNoogUc92mnjoxJwt8R166tx0=7JX04qFks_+g at mail dot gmail dot com>
On Thu, Mar 26, 2015 at 9:27 AM, Aleksey Midenkov <midenok@gmail.com> wrote:
> F.ex.
>
> define logging
> if $argc == 1
> if $arg0 == off
> set logging off
> set logging file gdb.log
> else if $arg0 == stop
> set logging off
> else
> set logging $arg0
> end
> else
> set logging $arg0 $arg1
> end
> show logging
> end
>
> When type 'logging off' I get error about no such symbol 'off'...
Such things are not supported in gdb's own scripting language.
However, with a bit of Python-provided magic ($_streq):
define logging
if $argc == 1
if $_streq("$arg0", "off")
set logging off
set logging file gdb.log
else
if $_streq("$arg0", "stop")
set logging off
else
set logging $arg0
end
end
else
set logging $arg0 $arg1
end
show logging
end
Note that gdb's if/else syntax is a pain.