Implicit constructor argument conversion in C++11

Question

Lets concider following code:

class A{
public:
  A(int x){}
};

class B{
public:
  B(A a){};
};


int main() {
  B b = 5;
  return 0;
}

And whi

Answer 1

You can use a converting constructor that is constrained on conversions to A.

class B {
public:
    // It doesn't hurt to keep that one
    B(A a){};

    template<
        typename T
        , EnableIf>...
    >
    B(T&& value)
    {
        // How to use the value
        A a = std::forward(value);
    }
};

// Now B b = foo; is valid iff A a = foo; is, except for a few exceptions

Make sure you understand the purpose of the EnableIf constraint. If you do not use it, you will face gnarly compilation errors, or worse: no errors at all. You should also carefully consider if making B convertible from potentially lots of types is at all worth it. Implicit conversions tend to make a program harder to understand, and that can quickly outweigh the apparent benefits they give.