Comment by svat

Yeah that's one of the things the test was designed to catch: at that point, B is a reference, to the B that is being defined. Here's a C++ translation from https://oeis.org/A132343 that uses identity functions to make the types consistent:

    #include <functional>
    #include <iostream>
    using cf = std::function<int()>;
    int A(int k, cf x1, cf x2, cf x3, cf x4, cf x5)
    {
        int Aval;
        cf B = [&]()
        {
            int Bval;
            --k;
            Bval = Aval = A(k, B, x1, x2, x3, x4);
            return Bval;
        };
        if (k <= 0) Aval = x4() + x5(); else B();
        return Aval;
    }
    cf I(int n) { return [=](){ return n; }; }
    int main()
    {
        for (int n=0; n<10; ++n)
            std::cout << A(n, I(1), I(-1), I(-1), I(1), I(0)) << ", ";
        std::cout << std::endl;
    }

So in the expression `A(k, B, x1, x2, x3, x4)`, the `B` there is not called, it simply refers to the local variable `B` (inside the function `A`), that was captured by the lambda (by reference): the same B variable that is currently being assigned.