How to transfer a string to the GPU and return it from the kernel in order to display it? (C ++ OpenCL)

Question

I would like to transfer the string to my GPU and return it from the GPU to print it. This is for understanding goals - I know the idea seems pointless.

I tried:

OpenCL:

__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; } 

But I failed. Output signal

Output: 5783885

Output [1..n]: 5783885 0 26

But not

MAX or 77 65 88

(for MAX).

Regard, Marcus

Answer 1

This is a fact giving you the answer you expect, but you are putting it in the wrong data type.

What you get is the only integer 5783885 , which is 0x0058414D (in hexadecimal format). You are on the little-endian platform, so these bytes are located in memory from low to high order, i.e. if you look at the memory, the bytes will be (still in hexadecimal format):

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

These are displayed in decimal form:

 77, 65, 88, 0, ... 

In other words, exactly what you expect.

Your problem (at least one of your problems) is that you declared out_global as an array of cl_uint s, not cl_uchar or cl_char or something like that.

So, change the following line and you will probably be fine.

 cl_uint out_global[bufferC_size];