PyCUDA: syntax for function that calls a function -

when using function sourcemodule depends on function in sourcemodule, how pass in function call, i.e. "???" in following code:

import numpy import pycuda.autoinit import pycuda.driver drv pycuda.compiler import sourcemodule  mod = sourcemodule(""" __global__ void make_square(float *in_array, float *out_array) {   int i;   int n = 5;   (i=0; i<n; i++)   {     out_array[i] = pow(in_array[i],2);   } } __global__ void make_square_add_one(float *in_array, float *out_array, void make_square(float *, float *)) {   int n = 5;   make_square(in_array,out_array);   (int i=0; i<n; i++)     out_array[i] = out_array[i] + 1; } """)  make_square = mod.get_function("make_square") make_square_add_one = mod.get_function("make_square_add_one") in_array = numpy.array([1.,2.,3.,4.,5.]).astype(numpy.float32) out_array = numpy.zeros_like(in_array).astype(numpy.float32) make_square_add_one(drv.in(in_array), drv.out(out_array), ??? , block = (1,1,1), grid = (1,1)) 

thanks information.

in traditional cuda execution model, __global__ functions kernels, , can't passed arguments other kernels , can't called other kernels. looks make_square should device function, like:

__device__ void make_square(float *in_array, float *out_array) {   int i;   (i=0; i<5; i++)   {     out_array[i] = pow(in_array[i],2);   } } 

which called running kernel as:

__global__ void make_square_add_one(float *in_array, float *out_array) {   int n = 5;   make_square(in_array,out_array);   (int i=0; i<n; i++)     out_array[i] = out_array[i] + 1; } 

it worth noting kernel entirely serial , pretty orthogonal how cuda kernels expected written.