1、cuda的学习记录Chapter01

基本思想：学习一下cuda编程，随手记录一下 如何使用grind block thread 线程块和编号的概念和应用方法

一、使用工具clion+ubuntu

CMakeLists.txt

cmake_minimum_required(VERSION 3.0)
project(untitled18)

set(CMAKE_CXX_STANDARD 14)
find_package(CUDA REQUIRED)
include_directories(${CUDA_INCLUDE_DIRS})
cuda_add_executable(untitled18 main.cu)

main.cu

#include <iostream>
#include <stdio.h>
#include "cuda.h"

__global__ void hello_from_gpu()
{

    int x=blockDim.x*blockIdx.x+threadIdx.x;
    printf("blockIdx.x %d threadIdx.x %d  x %d\n",blockIdx.x,threadIdx.x,x);
    //printf(" %d \n",x);
    //printf(" hello world\n");
}

int main() {
    hello_from_gpu<<<2,9>>>();// grid_dim block_dim  ===> block_id thread_id
    cudaDeviceSynchronize();
     return 0;
}

测试结果

/home/ubuntu/CLionProjects/untitled18/cmake-build-debug/untitled18
blockIdx.x 0 threadIdx.x 0  x 0
blockIdx.x 0 threadIdx.x 1  x 1
blockIdx.x 0 threadIdx.x 2  x 2
blockIdx.x 0 threadIdx.x 3  x 3
blockIdx.x 0 threadIdx.x 4  x 4
blockIdx.x 0 threadIdx.x 5  x 5
blockIdx.x 0 threadIdx.x 6  x 6
blockIdx.x 0 threadIdx.x 7  x 7
blockIdx.x 0 threadIdx.x 8  x 8
blockIdx.x 1 threadIdx.x 0  x 9
blockIdx.x 1 threadIdx.x 1  x 10
blockIdx.x 1 threadIdx.x 2  x 11
blockIdx.x 1 threadIdx.x 3  x 12
blockIdx.x 1 threadIdx.x 4  x 13
blockIdx.x 1 threadIdx.x 5  x 14
blockIdx.x 1 threadIdx.x 6  x 15
blockIdx.x 1 threadIdx.x 7  x 16
blockIdx.x 1 threadIdx.x 8  x 17

Process finished with exit code 0

二、测试程序

#include <iostream>
#include <stdio.h>
#include "cuda.h"
#include "algorithm"
#include "chrono"

__global__ void hello_from_gpu(int *aa,int *bb,int *cc)
{

    int index=blockDim.x*blockIdx.x+threadIdx.x;
    cc[index]=aa[index]+bb[index];
    printf("blockIdx.x %d threadIdx.x %d  index %d  \n",blockIdx.x,threadIdx.x,index);


}

int main() {
    int n=10;
    int *a=(int *)malloc(sizeof(int)*n);
    for(int i=0;i<n;i++){
        a[i]=2*i;

    }
    int *b=(int *)malloc(sizeof(int)*n);
    for(int i=0;i<n;i++){
        b[i]=i;

    }
    int *c=(int *)malloc(sizeof(int)*n);


    int *aa,*bb,*cc;

    cudaMalloc((void **)&aa,sizeof(int)*n);
    cudaMalloc((void **)&bb,sizeof(int)*n);
    cudaMalloc((void **)&cc,sizeof(int)*n);
    cudaMemcpy(aa,a,sizeof(int)*n,cudaMemcpyHostToDevice);
    cudaMemcpy(bb,b,sizeof(int)*n,cudaMemcpyHostToDevice);
    auto t1 = std::chrono::system_clock::now();

    hello_from_gpu<<<2,5>>>(aa,bb,cc);// grid_dim block_dim  ===> block_id thread_id
    cudaDeviceSynchronize();
    auto t2 = std::chrono::system_clock::now();
    auto elapsed = std::chrono::duration_cast<std::chrono::nanoseconds>(t2 - t1);

    std::cout << "Time difference: " << elapsed.count() << " ns" << std::endl;


    cudaMemcpy(c,cc,sizeof(int)*n,cudaMemcpyDeviceToHost);

    for(int i=0;i<n;i++){
        printf("%d+%d=%d\n",a[i],b[i],c[i]);
    }
    free(a);
    free(b);
    free(c);
    cudaFree(aa);
    cudaFree(bb);
    cudaFree(cc);
    return 0;
}

测试结果

/home/ubuntu/CLionProjects/untitled18/cmake-build-debug/untitled18
blockIdx.x 0 threadIdx.x 0  index 0  
blockIdx.x 0 threadIdx.x 1  index 1  
blockIdx.x 0 threadIdx.x 2  index 2  
blockIdx.x 0 threadIdx.x 3  index 3  
blockIdx.x 0 threadIdx.x 4  index 4  
blockIdx.x 1 threadIdx.x 0  index 5  
blockIdx.x 1 threadIdx.x 1  index 6  
blockIdx.x 1 threadIdx.x 2  index 7  
blockIdx.x 1 threadIdx.x 3  index 8  
blockIdx.x 1 threadIdx.x 4  index 9  
Time difference: 79481 ns
0+0=0
2+1=3
4+2=6
6+3=9
8+4=12
10+5=15
12+6=18
14+7=21
16+8=24
18+9=27

Process finished with exit code 0