On 04/15/2010 12:28 PM, Gleb Natapov wrote:
kvm_task_switch() never requires userspace exit, so no matter what the function returns we should not exit to userspace. Signed-off-by: Gleb Natapov<gleb@xxxxxxxxxx> diff --git a/arch/x86/kvm/svm.c b/arch/x86/kvm/svm.c index c773a46..1bd434b 100644 --- a/arch/x86/kvm/svm.c +++ b/arch/x86/kvm/svm.c @@ -2254,7 +2254,8 @@ static int task_switch_interception(struct vcpu_svm *svm) (int_vec == OF_VECTOR || int_vec == BP_VECTOR))) skip_emulated_instruction(&svm->vcpu); - return kvm_task_switch(&svm->vcpu, tss_selector, reason); + kvm_task_switch(&svm->vcpu, tss_selector, reason); + return 1; } static int cpuid_interception(struct vcpu_svm *svm) diff --git a/arch/x86/kvm/vmx.c b/arch/x86/kvm/vmx.c index 453f080..3e1607d 100644 --- a/arch/x86/kvm/vmx.c +++ b/arch/x86/kvm/vmx.c @@ -3306,8 +3306,7 @@ static int handle_task_switch(struct kvm_vcpu *vcpu) type != INTR_TYPE_NMI_INTR)) skip_emulated_instruction(vcpu); - if (!kvm_task_switch(vcpu, tss_selector, reason)) - return 0; + kvm_task_switch(vcpu, tss_selector, reason);
Ignoring the return seems wrong. At the very least log errors in dmesg (rate-limited).
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