C++ OpenCL:clSetKernelArg返回CL\u无效参数\u大小

我是OpenCL的新手。我试图将5个参数传递到内核中：一个输入缓冲区、一个输出缓冲区、一个整数和两个输入缓冲区大小的本地数组

//Create input/output cl_mem objects
cl_mem inputBuffer = clCreateBuffer(context, CL_MEM_READ_ONLY |
    CL_MEM_COPY_HOST_PTR, inputVector.size(), (void *)inputVector.data(), NULL);
cl_mem outputBuffer = clCreateBuffer(context, CL_MEM_WRITE_ONLY,
    inputVector.size(), NULL, NULL);

//get iterator from command line
cl_uint iterations = (cl_uint) atoi(argv[3]);
//std::cout << "iterations: " << iterations << std::endl
//cout confirms I'm getting this correctly

//Set kernel arguments
bool argOK = (CL_SUCCESS == clSetKernelArg(kernel, 0, sizeof(cl_mem), (void*)&inputBuffer));
argOK = argOK && (CL_SUCCESS == clSetKernelArg(kernel, 1, sizeof(cl_mem), (void *)&outputBuffer));
argOK = argOK && (CL_SUCCESS == clSetKernelArg(kernel, 2, sizeof(cl_uint), &iterations));
argOK = argOK && (CL_SUCCESS == clSetKernelArg(kernel, 3, sizeof(inputBuffer), NULL));
argOK = argOK && (CL_SUCCESS == clSetKernelArg(kernel, 4, sizeof(inputBuffer), NULL));
//Check for failure
if(!argOK) {
    std::cerr << "Error: clSetKernelArg failed\n";
    return SDK_FAILURE;
}

我做了一些挖掘/调试，最终发现这一行：

argOK = argOK && (CL_SUCCESS == clSetKernelArg(kernel, 2, sizeof(cl_uint), &iterations));

他是罪魁祸首。将其更改为：

argOK = argOK && (CL_INVALID_ARG_SIZE == clSetKernelArg(kernel, 2, sizeof(cl_uint), &iterations));

让程序成功继续因此，clSetKernelArg（kernel，2，…）语句返回CL_INVALID_ARG_SIZE。这很奇怪，因为看起来我传递了迭代变量的正确大小

以下是我的内核供参考：

__kernel void do_the_thing (__global uchar* in, __global uchar* out,
    __global uint * numIterations, __local uchar* arr1, __local uchar* arr2)
{
    //Do stuff that I haven't written yet.  

}

---

TL；DR:为什么setKernelArg调用返回CL\u无效的参数大小？内核的第三个参数（index=2）是缓冲区（

\uu全局uint*

）而不是标量

将内核更改为：

__kernel void do_the_thing (__global uchar* in, __global uchar* out, uint numIterations, __local uchar* arr1, __local uchar* arr2)
{
     //Do stuff that I haven't written yet.  

}

---

内核的第三个参数（index=2）是缓冲区（

\uu全局uint*

），而不是标量

将内核更改为：

__kernel void do_the_thing (__global uchar* in, __global uchar* out, uint numIterations, __local uchar* arr1, __local uchar* arr2)
{
     //Do stuff that I haven't written yet.  

}

---

这似乎可以做到，谢谢！所以uu global*意味着它是一个缓冲区，而把它像一个普通的c函数一样放进去，使它成为一个参数，那么？好吧，它不仅仅是

\uu global

。这同样适用于

\u常量

。似乎就是这样，谢谢！所以uu global*意味着它是一个缓冲区，而把它像一个普通的c函数一样放进去，使它成为一个参数，那么？好吧，它不仅仅是

\uu global

。这同样适用于

\u常量

。