On exposing intermediate measurement from kernel builder

[LennyDZ]

Hello,

If I build a kernel like :

@cudaq.kernel
def kernel():
    q = cudaq.qvector(2)
    reset(q[0])
    x(q[0])
    b0 = mz(q[0])
    if b0:
        h(q[1])
    mz(q[0])
    mz(q[1])
print(cudaq.sample(kernel))

Then b0 will be included in the SampleResult dictionary output.

But If I use kernel builder like :

k = cudaq.make_kernel()
q0 = k.qalloc()
q1 = k.qalloc()
k.reset(q0)
k.x(q0)
b0 = k.mz(q0) 
k.c_if(b0, lambda: k.h(q1))
k.mz(q0)
k.mz(q1)
o=cudaq.sample(k)
print(o)

Here b0isn't exposed in the output, I have to explicitly assign a register name to it like b0 = k.mz(q0, "b0") .
Not a big deal, but this isn't very intuitive. Is there a better way to do this ?

[efratshabtai]

Hi @LennyDZ, no, there is no better way to do that in builder mode. In builder mode, the compiler doesn't reason about Python code, you need to explicitly specify a name. We recommend using the kernel mode above.