如何将字符串传递给GPU并从内核返回以显示它？(C++ OpenCL)

Question

我想将一个字符串传递给我的GPU并从GPU中取回来打印它.这是为了理解目的 - 我知道,这个想法听起来毫无意义.

我试过了:

OpenCL的:

__kernel void same_in_same_out_char(__global uchar * out, __constant uchar * in){
  for (unsigned int ui=0; ui<3; ui++) out[ui]=in[ui];
}

C++:

#define __CL_ENABLE_EXCEPTIONS

#include <fstream>
#include <iostream>
#include <iterator>
#include <CL/cl.hpp>
#include <CL/opencl.h>


using namespace std;
int main () {

    vector<cl::Platform> platforms;
    vector<cl::Device> devices;
    vector<cl::Kernel> kernels;

    try {

        // create platform
        cl::Platform::get(&platforms);
        platforms[0].getDevices(CL_DEVICE_TYPE_GPU, &devices);

        // create context
        cl::Context context(devices);

        // create command queue
        cl::CommandQueue queue(context, devices[0]);

        // load opencl source
        ifstream cl_file("inout.cl");
        string cl_string(istreambuf_iterator<char>(cl_file), (istreambuf_iterator<char>()));
        cl::Program::Sources source(1, make_pair(cl_string.c_str(), 
            cl_string.length() + 1));

        // create program
        cl::Program program(context, source);

        // compile opencl source
        program.build(devices);

        // load named kernel from opencl source
        cl::Kernel kernel(program, "same_in_same_out_char");

        // create a message to send to kernel 
        const char pwd[] = "MAX";
        cout << "char pwd[] : " << pwd << endl;
        cl_uchar * password = (cl_uchar*) &pwd;
        int bufferA_size = 3; // array size is 3

        int bufferC_size = 3; // array size is 3
        cout << " -- OpenCL -- " << endl;

        // allocate device buffer to hold message
        cl::Buffer bufferA(context, CL_MEM_READ_ONLY | CL_MEM_COPY_HOST_PTR, sizeof(cl_uchar) * bufferA_size, password);
        cl::Buffer bufferC(context, CL_MEM_WRITE_ONLY, sizeof(cl_uchar) * bufferC_size);

        // set message as kernel argument
        kernel.setArg(0, bufferC);
        kernel.setArg(1, bufferA);

         // execute kernel
        queue.enqueueTask(kernel);

        // wait for completion
        queue.finish();
        // ----------------------

        cl_uint out_global[bufferC_size];
    queue.enqueueReadBuffer(bufferC, CL_TRUE, 0, bufferC_size*sizeof(cl_uchar), &out_global);

        cout << "Output \t\t:" << *out_global << endl << "Output[1..n] \t:";
        for (unsigned int i=0; i<bufferC_size; i ++) cout << out_global[i] << " " ;

        cout << endl;
    } catch (cl::Error e) {
        cout << endl << e.what() << " : " << e.err() << endl;
    }

    return 0;

}

但我失败了.输出是

产量:5783885

输出[1..n]:5783885 0 26

但不是

MAX或77 65 88

(对于MAX).

关注,马库斯

Answer 1

如果事实上给你你想要的答案,但你把它放在错误的数据类型.

你得到的是单个整数5783885,它是0x0058414D(十六进制).你是在一个小端平台上,所以这些字节在内存中从低位到高位排列,即如果你看看内存,字节将是(仍为十六进制):

4D, 41, 58, 00, ...

以十进制显示的这些将是:

77, 65, 88, 0, ...

换句话说,正是你所期望的.

您的问题(你的问题之一,至少)是你已经声明out_global为数组cl_uintS,而不是cl_uchar或cl_char什么的.

所以改变以下行,你可能会没事的.

cl_uint out_global[bufferC_size];