Re: how does ld.so call ELF's entry?

With a far jump to entry offset 


Sent from my iPad

On Apr 6, 2013, at 8:39 PM, ishare <[email protected]> wrote:

> On Sun, Apr 07, 2013 at 05:11:28AM +0200, Sofiane Akermoun wrote:
>> Hello,
>> 
>> 
>> The interpreter is specified during linking process and the program header
>> of your binary is filled with good values.
>> Then when the operating system load your binary, he finds next the
>> interpreter to use.
>> The linker ld sets the good values by default but you can overwrite it or
>> specify other values if you want.
>> The steps are:
>> 1)The operating system loads your binary
>> 2)The program loader system execute the Interpreter specifiy in the binary
>> 3)The interpreter gather all the dynamic libraries needed in memory
> 
> 
>> 4)The Control is passed to the entry point of your program
> 
>  How does  the step 4 been  done ?
> 
>  thanks!
>> 
>> The entry point is specified in your code source as a "global". And could
>> be find in your object file by the linker.
>> In theory there are some default tag to specify entry point, like "..start"
>> for nasm, but you can also passed it to the linker.
>> 
>> regards,
>> 
>> Sofiane Akermoun
>> [email protected]
>> 
>> 
>> 
>> 
>> 
>> 2013/4/7 ishare <[email protected]>
>> 
>>> 
>>>  For an ELF ,which needs a interpreter , how is it  been called by the
>>> interpreter ?
>>> 
>>>  As I know the interpreter is loaded first and do something essential
>>> ,then call the main routine of ELF .
>>>  How is this procedure implemented ?
>>> 
>>>  Thanks!
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>> 
>> 
>> 
>> -- 
>> Sofiane AKERMOUN
>> [email protected]
> --
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