CUDA, using memset(or fill or ...) to set an array of float to max val possible

Question

Edit: Thanks for the previous answers. but in fact I want to do it in CUDA, and apparently there is no function Fill for CUDA. I have to fill the matrix once for each thread so I want to make sure I'm using the fastest way possible. Is this for loop my best choice?

I want to set the matrix of float to the maximum value possible (in float). What is the correct way of doing this job?

float *matrix=new float[N*N];

for (int i=0;i<N*N;i++){
        matrix[i*N+j]=999999;
}

Thanks in advance.

Answer 1

The easiest approach in CUDA is to use thrust::fill. Thrust is included with CUDA 4.0 and later, or you can install it if you are using CUDA 3.2.

#include <thrust/fill.h>
#include <thrust/device_vector.h>
...
thrust::device_vector<float> v(N*N);
thrust::fill(v.begin(), v.end(), std::numeric_limits<float>::max()); // or 999999.f if you prefer

You could also write pure CUDA code something like this:

template <typename T>
__global__ void initMatrix(T *matrix, int width, int height, T val) {
    int idx = blockIdx.x * blockDim.x + threadIdx.x;

    for (int i = idx; i < width * height; i += gridDim.x * blockDim.x) {
        matrix[i]=val;
    }
}

int main(void) {
    float *matrix = 0;
    cudaMalloc((void*)&matrix, N*N * sizeof(float));

    int blockSize = 256;
    int numBlocks = (N*N + blockSize - 1) / (N*N);
    initMatrix<<<numBlocks, blockSize>>>(matrix, N, N, 
                                         std::numeric_limits<float>::max()); // or 999999.f if you prefer
}

Answer 2

Use std::numeric_limits<float>::max() and std::fill as:

#include <limits>     //for std::numeric_limits<> 
#include <algorithm>  //for std::fill

std::fill(matrix, matrix + N*N, std::numeric_limits<float>::max());

Or, std::fill_n as (looks better):

std::fill_n(matrix, N*N, std::numeric_limits<float>::max());

See these online documentation:

• std::fill
• std::fill_n