在不同线程上读取相同的内存位置时出错 [英] Error while reading same mem positions on different threads

问题描述

从不同的线程读取双精度数组中的几个位置时遇到问题.

我将执行加入队列:

nelements = nx*ny;
err = clEnqueueNDRangeKernel(queue,kernelTvl2of,1,NULL,&nelements,NULL,0,NULL,NULL);

kernelTvl2of具有(除其他外)代码

size_t k = get_global_id(0);
(...)
u1_[k] = (float)u1[k];
(...)
barrier(CLK_GLOBAL_MEM_FENCE);
forwardgradient(u1_,u1x,u1y,k,nx,ny);
barrier(CLK_GLOBAL_MEM_FENCE);

和正向渐变具有代码:

void forwardgradient(global double *f, global double *fx, global double *fy, int ker,int nx, int ny){
unsigned int rowsnotlast = ((nx)*(ny-1));
if(ker<rowsnotlast){
 fx[ker] = f[ker+1] - f[ker];
 fy[ker] = f[ker+nx] - f[ker];
}
if(ker<nx*ny){
 fx[ker] = f[ker+1] - f[ker];
 if(ker==4607){
  fx[0] = f[4607];
  fx[1] = f[4608];
  fx[2] = f[4608] - f[4607];
  fx[3] = f[ker];
  fx[4] = f[ker+1];
  fx[5] = f[ker+1] - f[ker];
 }
}
if(ker==(nx*ny)-1){
 fx[ker] = 0;
 fy[ker] = 0;
}
if(ker%nx == nx-1){
 fx[ker]=0;
}
fx[6] = f[4608];
}

当我得到fx的第一个位置的内容时，它们是:

-6 0 6 -6 0 6 -6

这是我的问题:当我在ID为4607的线程上查询fx [ker + 1]或fx [4608]时，得到一个"0"(输出数组的第二和第五位)，但是从其他线程得到输出数组的最后一个"-6"位置

任何人都知道我在做什么错，或者我会去哪里找?

非常感谢

安东

解决方案

在内核中，全局内存一致性只能在单个工作组中实现.这意味着，如果工作项将值写入全局内存，则barrier(CLK_GLOBAL_MEM_FENCE)仅保证相同工作组中的其他工作项将能够读取更新的值.

如果需要跨多个工作组的全局内存一致性，则需要将内核拆分为多个内核.

I have a problem while reading a couple of positions in a double array from different threads.

I enqueue the execution with :

nelements = nx*ny;
err = clEnqueueNDRangeKernel(queue,kernelTvl2of,1,NULL,&nelements,NULL,0,NULL,NULL);

kernelTvl2of has (among other) the code

size_t k = get_global_id(0);
(...)
u1_[k] = (float)u1[k];
(...)
barrier(CLK_GLOBAL_MEM_FENCE);
forwardgradient(u1_,u1x,u1y,k,nx,ny);
barrier(CLK_GLOBAL_MEM_FENCE);

and forwardgradient has the code:

void forwardgradient(global double *f, global double *fx, global double *fy, int ker,int nx, int ny){
unsigned int rowsnotlast = ((nx)*(ny-1));
if(ker<rowsnotlast){
 fx[ker] = f[ker+1] - f[ker];
 fy[ker] = f[ker+nx] - f[ker];
}
if(ker<nx*ny){
 fx[ker] = f[ker+1] - f[ker];
 if(ker==4607){
  fx[0] = f[4607];
  fx[1] = f[4608];
  fx[2] = f[4608] - f[4607];
  fx[3] = f[ker];
  fx[4] = f[ker+1];
  fx[5] = f[ker+1] - f[ker];
 }
}
if(ker==(nx*ny)-1){
 fx[ker] = 0;
 fy[ker] = 0;
}
if(ker%nx == nx-1){
 fx[ker]=0;
}
fx[6] = f[4608];
}

When I get the contents of the first positions of fx, they are:

-6 0 6 -6 0 6 -6

And here's my problem: when I query fx[ker+1] or fx[4608] on thread with id 4607 I get a '0' (positions second and fifth of the output array), but from other threads I get a '-6' last position of the output array)

Anyone has a clue on what I'm doing wrong, or where I could look to?

Thanks a lot,

Anton

解决方案

Within a kernel, global memory consistency is only achievable within a single work-group. This means that if a work-item writes a value to global memory, a barrier(CLK_GLOBAL_MEM_FENCE) only guarantees that other work-items within the same work-group will be able to read the updated value.

If you need global memory consistency across multiple work-groups, you need to split your kernel into multiple kernels.

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