Using C++ base class constructors?

Yes, Since C++11:

struct B2 {
    B2(int = 13, int = 42);
};
struct D2 : B2 {
    using B2::B2;
// The set of inherited constructors is
// 1. B2(const B2&)
// 2. B2(B2&&)
// 3. B2(int = 13, int = 42)
// 4. B2(int = 13)
// 5. B2()

// D2 has the following constructors:
// 1. D2()
// 2. D2(const D2&)
// 3. D2(D2&&)
// 4. D2(int, int) <- inherited
// 5. D2(int) <- inherited
};

For additional information see http://en.cppreference.com/w/cpp/language/using_declaration

Prefer initialization:

class C : public A
{
public:
    C(const string &val) : A(anInt) {}
};

In C++11, you can use inheriting constructors (which has the syntax seen in your example D).

Update: Inheriting Constructors have been available in GCC since version 4.8.

If you don't find initialization appealing (e.g. due to the number of possibilities in your actual case), then you might favor this approach for some TMP constructs:

class A
{
public: 
    A() {}
    virtual ~A() {}
    void init(int) { std::cout << "A\n"; }
};

class B : public A
{
public:
    B() : A() {}
    void init(int) { std::cout << "B\n"; }
};

class C : public A
{
public:
    C() : A() {}
    void init(int) { std::cout << "C\n"; }
};

class D : public A
{
public:
    D() : A() {}
    using A::init;
    void init(const std::string& s) { std::cout << "D -> " << s << "\n"; }
};

int main()
{
    B b; b.init(10);
    C c; c.init(10);
    D d; d.init(10); d.init("a");

    return 0;
}