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On my OS X box, the kernel is a 32-bit binary and yet it can run a 64-bit binary. How does this work?
cristi:~ diciu$ file ./a.out ./a.out: Mach-O 64-bit executable x86_64 cristi:~ diciu$ file /mach_kernel /mach_kernel: Mach-O universal binary with 2 architectures /mach_kernel (for architecture i386): Mach-O executable i386 /mach_kernel (for architecture ppc): Mach-O executable ppc cristi:~ diciu$ ./a.out cristi:~ diciu$ echo $? 1
The CPU can be switched from 64 bit execution mode to 32 bit when it traps into kernel context, and a 32 bit kernel can still be constructed to understand the structures passed in from 64 bit user-space apps.
The MacOS X kernel does not directly dereference pointers from the user app anyway, as it resides its own separate address space. A user-space pointer in an ioctl call, for example, must first be resolved to its physical address and then a new virtual address created in the kernel address space. It doesn’t really matter whether that pointer in the ioctl was 64 bits or 32 bits, the kernel does not dereference it directly in either case.
So mixing a 32 bit kernel and 64 bit binaries can work, and vice-versa. The thing you cannot do is mix 32 bit libraries with a 64 bit application, as pointers passed between them would be truncated. MacOS X supplies more of its frameworks in both 32 and 64 bit versions in each release.