how to declare a double array in javascript for this particular UI

Hope this helps You don't have to define variable types in Matlab. To declare C and D, simply do the following:

C = zeros(240,500);
D = cell(1,500);

x = 100;
C = ones(240,500) * 100;

it helps some times Apparently the file on GitHub does use std::array, but doesn't include the array standard header.
Including it solved the issue.

wish of those help Assuming you're going to access the data in a block-oriented manner, you want to do the latter. This is presumably because when you load the first element of the "len" dimension, you've already paid the cost for missing in the cache for the subsequent 7ish elements. In the first option, there's probably sharing of cache lines between GPU blocks, but the sharing is relatively limited and not as low level.
Indeed, the below code reports that the second option requires 0.481 seconds to execute, and the first option requires 0.979 seconds. Arranging the data with the block on the outer dimension is about twice as performant.

#include <cuda_runtime_api.h>
#include <cuda.h>

#include <string>
#include <chrono>
#include <iostream>

#define BLOCKS 80
#define LEN (1 << 20)

void CheckCudaErrorAux (const char *file, unsigned line, const char *statement, cudaError_t err) {
    if (err == cudaSuccess)
        return;
    std::cerr << statement<<" returned " << cudaGetErrorString(err) << "("<<err<< ") at "<<file<<":"<<line << std::endl;
    exit (1);
}
#define CUDA_CHECK_RETURN(value) CheckCudaErrorAux(__FILE__,__LINE__, #value, value)

struct Data1 {
    int payload[LEN][BLOCKS];
};

struct Data2 {
    int payload[BLOCKS][LEN];
};


__global__ void f1(Data1 * data1) {
    int sum = 0;
    for (int i = 0; i < LEN; ++i) {
        sum += data1->payload[i][blockIdx.x];
    }
    printf("block %i has f1 sum %i\n", blockIdx.x, sum);
}

__global__ void f2(Data2 * data2) {
    int sum = 0;
    for (int i = 0; i < LEN; ++i) {
        sum += data2->payload[blockIdx.x][i];
    }
    printf("block %i has f2 sum %i\n", blockIdx.x, sum);
}


int main() {

    Data1 * data1 = (Data1 *) malloc(sizeof(Data1));
    Data2 * data2 = (Data2 *) malloc(sizeof(Data2));;

    for (int i = 0; i < LEN; ++i) {
        for (int b = 0; b < BLOCKS; ++b) {
            data1->payload[i][b] = i * b;
            data2->payload[b][i] = i * b;
        }
    }

    Data1 * data1_on_gpu;
    CUDA_CHECK_RETURN(cudaMalloc(&data1_on_gpu, sizeof(Data1)));
    Data2 * data2_on_gpu;
    cudaMalloc(&data2_on_gpu, sizeof(Data2));
    CUDA_CHECK_RETURN(cudaDeviceSynchronize());
    cudaMemcpy(data1_on_gpu, data1, sizeof(Data1), cudaMemcpyHostToDevice);
    cudaMemcpy(data2_on_gpu, data2, sizeof(Data1), cudaMemcpyHostToDevice);
    CUDA_CHECK_RETURN(cudaDeviceSynchronize());


    std::chrono::time_point<std::chrono::system_clock> t1 = std::chrono::system_clock::now();

    f1<<<80,1>>>(data1_on_gpu);
    CUDA_CHECK_RETURN(cudaDeviceSynchronize());
    std::chrono::time_point<std::chrono::system_clock> t2 = std::chrono::system_clock::now();

    f2<<<80,1>>>(data2_on_gpu);
    CUDA_CHECK_RETURN(cudaDeviceSynchronize());
    std::chrono::time_point<std::chrono::system_clock> t3 = std::chrono::system_clock::now();


    std::chrono::duration<double> duration_1_to_2 = t2 - t1;
    std::chrono::duration<double> duration_2_to_3 = t3 - t2;
    duration_1_to_2.count();

    printf("timer for 1st took %.3lf\n", duration_1_to_2.count());
    printf("timer for 2nd took %.3lf\n", duration_2_to_3.count());

}