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Re: ld -e number makes an entry in symbol table


On Thu, Sep 10, 2009 at 8:05 AM, Alan Modra<amodra@bigpond.net.au> wrote:
> On Thu, Sep 10, 2009 at 06:28:40AM -0700, H.J. Lu wrote:
>> On Wed, Sep 9, 2009 at 8:41 PM, Alan Modra<amodra@bigpond.net.au> wrote:
>> > On Wed, Sep 09, 2009 at 08:50:23AM -0700, H.J. Lu wrote:
>> >> On Wed, Sep 9, 2009 at 8:23 AM, Andreas Schwab<schwab@redhat.com> wrote:
>> >> > This was broken by the patch for PR ld/6766.
>> >>
>> >> Here is a patch. ?OK to install?
>> >
>> > I obviously gave the OK for your PR6766 patch too quickly. ?I'm
>> > inclined to think the whole patch should be reverted.
>> >
>> > Why should -e be any different from ENTRY() in a script?
>> >
>>
>> 1. You can only use symbol with ENTRY ().
>
> Yes, that's a difference but I meant in the context of your PR6766
> change.
>
>> 2. "-e" is specified by user as the start address.
>
> A user can supply a script with ENTRY.
>
>> 3. "-e" overrides ENTRY ().
>
> And -e overrides a previous -e.
>
> My question was regarding -e implying -u. ?If -e implies -u, then
> shouldn't ENTRY imply EXTERN in a script? ?If not, why not?
> Because the default scripts have ENTRY, and these scripts are not user
> input, but a command line switch is user input? ?Ah, but ld is often
> invoked by other tools, eg. gcc, passing all sorts of options that a
> naive user never sees. ?If gcc passes -e to ld is it still user input?
> What if some IDE passes -e to ld?
>

By "user", I meant the user of linker, like programmer, compiler, .....

When "-e SYMBOL" is passed to linker, SYMBOL should be used
as the entry point. If SYMBOL doesn't exist, something is wrong.
ENTRY is defined in the linker scripts, which may be independent
of the user.

-- 
H.J.


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