"""We don't have real test for the cache, but it would be great to make them!

But this one test a current behavior that isn't good: the c_code isn't
deterministic based on the input type and the op.

"""
import numpy

import theano


class MyOp(theano.compile.ops.DeepCopyOp):
    nb_called = 0

    def c_code_cache_version(self):
        return ()

    def c_code(self, node, name, inames, onames, sub):
        MyOp.nb_called += 1
        iname, = inames
        oname, = onames
        fail = sub['fail']
        itype = node.inputs[0].type.__class__
        if itype in self.c_code_and_version:
            code, version = self.c_code_and_version[itype]
            rand = numpy.random.rand()
            return ("""printf("%(rand)s\\n");""" + code) % locals()
        # Else, no C code
        return super(DeepCopyOp, self).c_code(node, name, inames, onames, sub)


def test_inter_process_cache():
    """When an op with c_code, but no version. If we have 2 apply node
    in the graph with different inputs variable(so they don't get
    merged) but the inputs variable have the same type, do we reuse
    the same module? Even if they would generate different c_code?
    Currently this test show that we generate the c_code only once.

    This is to know if the c_code can add information specific to the
    node.inputs[*].owner like the name of the variable.

    """

    x, y = theano.tensor.vectors('xy')
    f = theano.function([x, y], [MyOp()(x), MyOp()(y)])
    f(numpy.arange(60), numpy.arange(60))
    assert MyOp.nb_called == 1

    # What if we compile a new function with new variables?
    x, y = theano.tensor.vectors('xy')
    f = theano.function([x, y], [MyOp()(x), MyOp()(y)])
    f(numpy.arange(60), numpy.arange(60))
    assert MyOp.nb_called == 1
