Memory is always the destination for BitOp instructions. Signed-off-by: Paolo Bonzini <pbonzini@redhat.com>
@@ -4079,7 +4079,7 @@ static int decode_operand(struct x86_emulate_ctxt *ctxt, struct operand *op,
mem_common:
*op = ctxt->memop;
ctxt->memopp = op;
- if ((ctxt->d & BitOp) && op == &ctxt->dst)
+ if (ctxt->d & BitOp)
fetch_bit_operand(ctxt);
op->orig_val = op->val;
break;