Can empty kernels be stripped during translation to OpenQASM 2?

[kshyatt-aws]

CUDA-QUANTUM version: 0.8.0
OS: Amazon Linux 2023 x86_64

Looking the basic C++ GHZ script:

#include <cudaq.h>

template <std::size_t N>
struct ghz {
  auto operator()() __qpu__ {
    cudaq::qarray<N> q;
    h(q[0]);
    for (int i = 0; i < N - 1; i++) {
      x<cudaq::ctrl>(q[i], q[i + 1]);
    }
    mz(q);
  }
};

int main() {

  auto kernel = ghz<10>{};
  auto counts = cudaq::sample(kernel);

  if (!cudaq::mpi::is_initialized() || cudaq::mpi::rank() == 0) {
    counts.dump();

    // Fine grain access to the bits and counts
    for (auto &[bits, count] : counts) {
      printf("Observed: %s, %lu\n", bits.data(), count);
    }
  }

  return 0;
}

When I try to compile and translate this to OQ2 using:

cudaq-quake ghz.cpp | cudaq-opt --pass-pipeline="builtin.module(canonicalize,lambda-lifting,apply-op-specialization,func.func(memtoreg{quantum=0}),cc-loop-normalize,cc-loop-unroll)"  | cudaq-translate --convert-to=openqasm2 &> ghz.qasm

Here's the output qasm file:

// Code generated by NVIDIA's nvq++ compiler
OPENQASM 2.0;

include "qelib1.inc";

gate ZN3ghzILm10EEclEv(param0)  {
}

qreg var0[10];
h var0[0];
cx var0[0], var0[1];
cx var0[1], var0[2];
cx var0[2], var0[3];
cx var0[3], var0[4];
cx var0[4], var0[5];
cx var0[5], var0[6];
cx var0[6], var0[7];
cx var0[7], var0[8];
cx var0[8], var0[9];
creg var11[10];
measure var0 -> var11;

Looks good! But is there a way to strip the extraneous

gate ZN3ghzILm10EEclEv(param0)  {
}

during the cudaq-opt step? I looked at the cudaq-opt --help output but did not see an pipeline-pass for this, if one exists.

[schweitzpgi]

One idea that comes to mind is for the OpenQASM $n$ pretty-printer to skip the C++ entry points. (Perhaps under a command-line option, although those functions, though required in the grand scheme, really won't be interesting as far as OpenQASM $n$ and are not device code.)