Il 01/06/2014 01:05, Rickard Strandqvist ha scritto:
> There is a risk that the variable will be used without being initialized.
>
> This was largely found by using a static code analysis program called cppcheck.
>
> Signed-off-by: Rickard Strandqvist <rickard_strandqvist@xxxxxxxxxxxxxxxxxx>
No, there isn't. The full context looks like this:
longmode = is_long_mode(vcpu) && cs_l == 1;
if (!longmode) {
param = ((u64)kvm_register_read(vcpu, VCPU_REGS_RDX) << 32) |
(kvm_register_read(vcpu, VCPU_REGS_RAX) & 0xffffffff);
ingpa = ((u64)kvm_register_read(vcpu, VCPU_REGS_RBX) << 32) |
(kvm_register_read(vcpu, VCPU_REGS_RCX) & 0xffffffff);
outgpa = ((u64)kvm_register_read(vcpu, VCPU_REGS_RDI) << 32) |
(kvm_register_read(vcpu, VCPU_REGS_RSI) & 0xffffffff);
}
#ifdef CONFIG_X86_64
else {
param = kvm_register_read(vcpu, VCPU_REGS_RCX);
ingpa = kvm_register_read(vcpu, VCPU_REGS_RDX);
outgpa = kvm_register_read(vcpu, VCPU_REGS_R8);
}
#endif
and longmode must be zero if !CONFIG_X86_64:
static inline int is_long_mode(struct kvm_vcpu *vcpu)
{
#ifdef CONFIG_X86_64
return vcpu->arch.efer & EFER_LMA;
#else
return 0;
#endif
}
So it's the static checker that cannot understand #ifdef well
enough and ought to be fixed.
Paolo
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