Re: [Qemu-devel] [PATCH] don't expose lm bit if kernel is not 64-bit capable.

[Alexander Graf]

On 03.02.2009, at 16:04, Glauber Costa wrote:

> If the kernel is not 64-bit capable (even if the host
> machine is) do not expose the lm bit in guest cpuid.
>
> Signed-off-by: Glauber Costa <address@hidden>
> ---
> target-i386/helper.c |    6 +++++-
> 1 files changed, 5 insertions(+), 1 deletions(-)
>
> diff --git a/target-i386/helper.c b/target-i386/helper.c
> index a28ab93..997e4e1 100644
> --- a/target-i386/helper.c
> +++ b/target-i386/helper.c
> @@ -24,6 +24,7 @@
> #include <inttypes.h>
> #include <signal.h>
> #include <assert.h>
> +#include <sys/utsname.h>
>
> #include "cpu.h"
> #include "exec-all.h"
> @@ -1520,13 +1521,16 @@ void cpu_x86_cpuid(CPUX86State *env,  
> uint32_t index,
>
>         if (kvm_enabled()) {
>             uint32_t h_eax, h_edx;
> +            struct utsname utsname;
> +
> +            uname(&utsname);

This doesn't really compile on win32, does it?

Why not check for sizeof(long) == 4? That way you'd know that you're  
running 32-bit code which can only handle 32-bit guests anyway, right?  
And if you're running 64-bit userspace code, you know that the kernel  
is 64-bit aware.

Alex

>             host_cpuid(0x80000001, &h_eax, NULL, NULL, &h_edx);
>
>             /* disable CPU features that the host does not support */
>
>             /* long mode */
> -            if ((h_edx & 0x20000000) == 0 /* || !lm_capable_kernel  
> */)
> +            if ((h_edx & 0x20000000) == 0  ||  
> strcmp(utsname.machine, "x86_64"))
>                 *edx &= ~0x20000000;
>             /* syscall */
>             if ((h_edx & 0x00000800) == 0)
> --
> 1.5.6.5