MkBVSHL() публичный Метод

Shift left.
It is equivalent to multiplication by 2^x where \c x is the value of t2. NB. The semantics of shift operations varies between environments. This definition does not necessarily capture directly the semantics of the programming language or assembly architecture you are modeling. The arguments must have a bit-vector sort.
public MkBVSHL ( BitVecExpr t1, BitVecExpr t2 ) : BitVecExpr
t1 BitVecExpr
t2 BitVecExpr
Результат BitVecExpr
Документация по классу Context
Пример #1
0
    public void Run()
    {
        Dictionary<string, string> cfg = new Dictionary<string, string>() {
            { "AUTO_CONFIG", "true" } };

        using (Context ctx = new Context(cfg))
        {
            BitVecExpr x = ctx.MkBVConst("x", 32);
            BitVecExpr y = ctx.MkBVConst("y", 32);

            BitVecExpr two = ctx.MkBV(2, 32);
            BitVecExpr three = ctx.MkBV(3, 32);
            BitVecExpr tf = ctx.MkBV(24, 32);

            Solver s = ctx.MkSolver();
            s.Assert(ctx.MkEq(ctx.MkBVLSHR(x, two), three));
            Console.WriteLine(s.Check());
            Console.WriteLine(s.Model);

            s = ctx.MkSolver();
            s.Assert(ctx.MkEq(ctx.MkBVSHL(x, two), three));
            Console.WriteLine(s.Check());

            s = ctx.MkSolver();
            s.Assert(ctx.MkEq(ctx.MkBVRotateLeft(x, two), three));
            Console.WriteLine(s.Check());
            Console.WriteLine(s.Model);

            s = ctx.MkSolver();
            s.Assert(ctx.MkEq(ctx.MkBVSHL(x, two), tf));
            Console.WriteLine(s.Check());
            Console.WriteLine(s.Model);
        }
    }
Пример #2
0
    public void Run()
    {
        Dictionary<string, string> cfg = new Dictionary<string, string>() {
            { "AUTO_CONFIG", "true" } };

        using (Context ctx = new Context(cfg))
        {
            BitVecExpr x = ctx.MkBVConst("x", 32);
            BitVecExpr[] powers = new BitVecExpr[32];
            for (uint i = 0; i < 32; i++)
                powers[i] = ctx.MkBVSHL(ctx.MkBV(1, 32), ctx.MkBV(i, 32));

            BoolExpr step_zero = ctx.MkEq(ctx.MkBVAND(x, ctx.MkBVSub(x, ctx.MkBV(1, 32))), ctx.MkBV(0, 32));

            BoolExpr fast = ctx.MkAnd(ctx.MkNot(ctx.MkEq(x, ctx.MkBV(0, 32))),
                                      step_zero);

            BoolExpr slow = ctx.MkFalse();
            foreach (BitVecExpr p in powers)
                slow = ctx.MkOr(slow, ctx.MkEq(x, p));

        TestDriver.CheckString(fast, "(and (not (= x #x00000000)) (= (bvand x (bvsub x #x00000001)) #x00000000))");

            Solver s = ctx.MkSolver();
            s.Assert(ctx.MkNot(ctx.MkEq(fast, slow)));
            TestDriver.CheckUNSAT(s.Check());

            s = ctx.MkSolver();
            s.Assert(ctx.MkNot(step_zero));
            TestDriver.CheckSAT(s.Check());
        }
    }