I am a physicist getting acquainted with one of the typical constructs for formulation and analysis of quantum algorithms (such as search problems or query complexity models), namely the "oracle function". As far as I understand, the oracle is a quantum black-box entity which computes the quantity to be analyzed. For example, in a qubit circuit formulation of a universal quantum computer the oracle is a deterministically computed Boolean function (correct me here if I'm wrong).

My question is:

* Is quantum oracle implementation overhead counted when the
computational complexity of specific/well-known algorithms is compared to
alternatives?

**Examples**: deterministic [Deutsch-Jozsa algorithm](http://en.wikipedia.org/wiki/Deutsch%E2%80%93Jozsa_algorithm) algorithm is exponentially faster than a classical counterpart only if the implementation of the unknown function $f$ requires less than $2^n$ classical "steps'' (including obtaining the guarantee that $f$ is either *partial* or *constant*). A stochastic example: collapse of the Grover algorithm speedup by a [faulty oracle](http://www.cims.nyu.edu/~regev/papers/faultygrover.pdf) with an arbitrary small but constant failure probability.

**Why do one care**: Of course, one can alsways say that we look just at the non-oracle part or count only the number of queries etc. But any talk of quantum speedup has any chance of empirical relevance only if the overhead for implementation of all of the quantum parts of the construction is taken into account (if it is trivial/constant - no problem). In other words, I find it meaningful to compare quantum algorithms only after they are "fully compiled'' onto (at least hypothetically) realizable classically-driven hardware (e.g., universal set of gates plus as many qubits as one needs - but we'll count all the time & memory required). 

My question is probably fairly typical for a non-expert, perhaps I just got confused by some less sophisticated introductions to quantum computation. Please deconfuse me.