Is compat_sys_ioctl called when both kernel and userland are 64bit ?

[april]

Hi all:

I am doing something to support 32bit userland and 64bit kernel.
I know compat_sys_ioctl will be called if 32bit userland send ioctl to
64bit kernel.

The compat ioctl do some conversion(suppose I provide those ioctls).
For example:
if a struct has a field  type is long or a pointer,which the size is
different between userland and kernel

I notice in the x86_64_defconfig file:
CONFIG_X86_64=y
CONFIG_IA32_EMULATION=y
and in "trap_init" function,
#ifdef CONFIG_IA32_EMULATION
        set_system_intr_gate(IA32_SYSCALL_VECTOR, ia32_syscall);
        set_bit(IA32_SYSCALL_VECTOR, used_vectors);
#endif

it seems when a kernel is built to x86_64, whenever a ioctl called
,the call flow will be ia32_syscall -->compat_sys_ioctl->compat
ioctl(if provided)
but I have some doubt:

My question is:
1.  Is compat_sys_ioctl called when both kernel and userland are 64bit
(suppose I provide those compat ioctl functions)?
     if so, it may have problem, for the compat ioctl suppose userland
is 32bit, and do the conversion, if a field is unsinged long or a
pointer, it will do the        wrong things .

     If not, how kernel knows the driver is 32bit or 64bit?

2. when using mmap, the 64bit kernel will return a 64bit address, and
a userland(32bit) application can only get the lower 32bit,
it can work when the memory is not large.

Is kernel do some conversion? or fortunately the higher 32bit is 00000000 ?

Thanks
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