cython

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# mode: run
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# tag: assign, exttype
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cdef struct X:
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    int ix
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    X* x
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cdef class A:
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    cdef int i
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    cdef list l
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    cdef object o
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    cdef X x
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    def assign_A(self):
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        """
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        >>> A().assign_A()
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        (2, [1, 2, 3])
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        """
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        a = A()
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        a.i = 1
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        a.l = [1, 2, 3]
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        a.o = a.l
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        a.o = a.o
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        a.l = a.o
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        a.i = a.l[1]
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        return a.i, a.l
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    def assign_A_struct(self):
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        """
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        >>> A().assign_A_struct()
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        (5, 2, 2, 5)
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        """
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        cdef X x
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        a = A()
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        a.x.ix = 2
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        a.x.x = &x
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        x.ix = 5
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        x.x = &a.x
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        assert a.x.x.x is &a.x
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        a.x.x.x.x.x.x.x = a.x.x.x.x
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        assert x.x is &x
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        assert x.x.x is &x
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        assert a.x.x is &x
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        a.x.x.x.x.x.x.x, a.x.x.x = a.x.x.x.x, &a.x   # replay+undo :)
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        assert x.x is &a.x
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        assert x.x.x is &x
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        return x.ix, x.x.ix, a.x.ix, a.x.x.ix
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cdef class B(A):
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    cdef int ib
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    cdef object ob
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    cdef A a
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    def assign_B(self):
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        """
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        >>> B().assign_B()
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        (1, 2, 5, 9, 2)
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        """
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        b = B()
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        b.i = 1
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        b.ib = 2
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        b.l = [b.i, b.ib]
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        b.o = b.l
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        b.ob = b.o
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        assert b.ob == b.l
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        b.o = b.ob = b.l
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        b.a = A()   # only one reference!
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        b.a.o = 5
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        b.a.i = 5
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        b.a, b.a.i = A(), b.a.i  # overwrite b.a but keep b.a.i
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        assert b.a.i == 5
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        assert b.a.o is None
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        b.a.o = 9
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        b.a, b.a.i, b.a.o = A(), b.a.i, b.a.o
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        return b.i, b.ib, b.a.i, b.a.o, b.o[1]
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    def cross_assign_Ba(self):
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        """
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        >>> B().cross_assign_Ba()
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        2
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        """
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        b = B()
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        b.a = A()
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        b.a.i = 1
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        b.a.o = A()   # only one reference!
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        (<A>b.a.o).i = 2
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        b.a = b.a.o
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        return b.a.i
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    def cascaded_assign_B(self):
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        """
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        >>> B().cascaded_assign_B()
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        (2, 2)
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        """
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        cdef B b = B()
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        b.ib = 1
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        b.a = A()
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        b.a.o = B()   # only one reference!
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        (<B>b.a.o).ib = 2
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        b = b.ob = b.a.o
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        return b.ib, (<B>b.ob).ib
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