CUDA: Numerical differences between CPython / CPU target and the CUDA target in `math.pow`

Originally noticed on Discourse.

The following:

from numba import cuda, njit
import numpy as np
import math


def inner(x, y):
    return int(math.pow(x, y))


cpu_jitted = njit(inner)


def cuda_jitted(x, y):
    inner_device = cuda.jit(device=True)(inner)

    @cuda.jit
    def outer(r, x, y):
        r[()] = inner_device(x[()], y[()])

    r_arr = np.array(0, dtype=np.int64)
    x_arr = np.array(0, dtype=np.int64)
    y_arr = np.array(0, dtype=np.int64)
    x_arr[()] = x
    y_arr[()] = y
    outer[1, 1](r_arr, x_arr, y_arr)

    return r_arr[()]


x = 3
y = 1

print(inner(x, y))
print(cpu_jitted(x, y))
print(cuda_jitted(x, y))

prints

3
3
2

This can cause some surprise.

[gmarkall]

IIRC a similar issue affected cuDF: https://github.com/rapidsai/cudf/issues/10178

[bdice]

If the result is expected to be integral, it is possible to solve this in the same way as CuPy with an exponentiation-by-squaring algorithm, as I noted in the cuDF issue you linked. I think that might be the best solution for Numba as well as cuDF. https://github.com/rapidsai/cudf/issues/10178#issuecomment-1029200496