On March 7, 2016 12:22:28 AM PST, Ingo Molnar <mi...@kernel.org> wrote:
>
>* Andy Lutomirski <l...@kernel.org> wrote:
>
>> Ingo suggested that the comments should explain when the various
>> entries are used.  This adds these explanations and improves other
>> parts of the comments.
>
>Thanks for doing this, this is really useful!
>
>One very small detail I noticed:
>
>> +/*
>> + * 32-bit legacy system call entry.
>> + *
>> + * 32-bit x86 Linux system calls traditionally used the INT $0x80
>> + * instruction.  INT $0x80 lands here.
>> + *
>> + * This entry point can be used by 32-bit and 64-bit programs to
>perform
>> + * 32-bit system calls.  Instances of INT $0x80 can be found inline
>in
>> + * various programs and libraries.  It is also used by the vDSO's
>> + * __kernel_vsyscall fallback for hardware that doesn't support a
>faster
>> + * entry method.  Restarted 32-bit system calls also fall back to
>INT
>> + * $0x80 regardless of what instruction was originally used to do
>the
>> + * system call.
>> + *
>> + * This is considered a slow path.  It is not used by modern libc
>> + * implementations on modern hardware except during process startup.
>> + *
>> + * Arguments:
>> + * eax  system call number
>> + * ebx  arg1
>> + * ecx  arg2
>> + * edx  arg3
>> + * esi  arg4
>> + * edi  arg5
>> + * ebp  arg6
>> + */
>>  ENTRY(entry_INT80_32)
>
>entry_INT80_32() is only used on pure 32-bit kernels, 64-bit kernels
>use 
>entry_INT80_compat(). So the above text should not talk about 64-bit
>programs, as 
>they can never trigger this specific entry point, right?
>
>So I'd change the explanation to something like:
>
>> + * This entry point is active on 32-bit kernels and can thus be used
>by 32-bit 
>> + * programs to perform 32-bit system calls. (Programs running on
>64-bit
>> + * kernels executing INT $0x80 will land on another entry point: 
>> + * entry_INT80_compat. The ABI is identical.)
>
>Agreed?
>
>Thanks,
>
>       Ingo

Sadly I believe Android still uses int $0x80 in the upstream version.
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