Phase gate implementation using CUDA-Q. The example applies $P(\pi/2) = S$ to $\lvert +\rangle$, rotating it to $\lvert i\rangle = \frac{1}{\sqrt{2}}(\lvert 0\rangle + i\lvert 1\rangle)$.

// Compile: nvq++ main.cpp -o main
// Run:     ./main
 
#include <cudaq.h>
#include <cmath>
 
struct kernel {
    __qpu__ void operator()() {
        cudaq::qubit q;
        h(q);                    // |0> -> |+>
        r1(M_PI / 2.0, q);       // P(pi/2) = S gate
        mz(q);
    }
};
 
int main() {
    auto counts = cudaq::sample(kernel{});
    counts.dump();  // ~50% |0>, ~50% |1>  (phase not observable in Z basis)
}