Re: How to compare $arg0 with string literal?

[Doug Evans <dje@google.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.