How to square a set of two numbers?
In this lesson, we're going to take an input array of 64 floating point numbers, numbered 0 to 63, and we're going to square each number in the array. So the output will be 0, 1, 4, 9, and so on. We're going to do this in three steps.
- We're going to start by looking at how we'd run this code on the CPU.
- Then we'll talk about how we'd run this on the GPU without looking at code,
- Instead just discussing what our code would do. Then we'll dive into what the code actually looks like.
In the following, we run the code from the above video lesson.
#include <stdio.h>
__global__ void cube(float * d_out, float * d_in){
// Todo: Fill in this function
int idx = threadIdx.x;
float f = d_in[idx];
d_out[idx] = f*f;
}
int main(int argc, char ** argv) {
const int ARRAY_SIZE = 96;
const int ARRAY_BYTES = ARRAY_SIZE * sizeof(float);
// generate the input array on the host
float h_in[ARRAY_SIZE];
for (int i = 0; i < ARRAY_SIZE; i++) {
h_in[i] = float(i);
}
float h_out[ARRAY_SIZE];
// declare GPU memory pointers
float * d_in;
float * d_out;
// allocate GPU memory
cudaMalloc((void**) &d_in, ARRAY_BYTES);
cudaMalloc((void**) &d_out, ARRAY_BYTES);
// transfer the array to the GPU
cudaMemcpy(d_in, h_in, ARRAY_BYTES, cudaMemcpyHostToDevice);
// launch the kernel
cube<<<1, 128>>>(d_out, d_in);
// copy back the result array to the CPU
cudaMemcpy(h_out, d_out, ARRAY_BYTES, cudaMemcpyDeviceToHost);
// print out the resulting array
for (int i =0; i < ARRAY_SIZE; i++) {
printf("%f", h_out[i]);
printf(((i % 4) != 3) ? "\t" : "\n");
}
cudaFree(d_in);
cudaFree(d_out);
return 0;
}c++