atomicInc（）不工作 [英] atomicInc() is not working

问题描述

我试过下面的程序使用atomicInc（）。

  __ global__ void ker（int * count）
 { 
 int n = 1; 
 int x = atomicInc（（unsigned int *）& count [0]，n）; 
 CUPRINTF（在内核计数为％d \\\
，count [0]）; 
} 
 
 int main（）
 {
 int hitCount [1]; 
 int * hitCount_d; 
 
 hitCount [0] = 1; 
 cudaMalloc（（void **）& hitCount_d，1 * sizeof（int））; 
 
 cudaMemcpy（& hitCount_d [0]，& hitCount [0]，1 * sizeof（int），cudaMemcpyHostToDevice）; 
 
 ker<<< 1,4>>>（hitCount_d）; 
 
 cudaMemcpy（& hitCount [0]，& hitCount_d [0]，1 * sizeof（int），cudaMemcpyDeviceToHost）; 
 
 printf（count is％d\\\
，hitCount [0]）; 
 return 0; 
} 
  

输出为：

 内核计数为1 
内核计数为1 
内核计数为1 
内核计数为1 
 
 count是1 
  

我不明白为什么它不递增。任何人都可以帮助

解决方案

引用文档， atomicInc 执行此操作：

：

  atomicInc（unsigned int *）& count [0] n）。 
  

计算：

 code>（（count [0]> = n）？0：（count [0] +1））
  

并将结果存回 count [0]

不确定？操作符是什么，请查看这里）

由于您已经通过 n = 1和计数[0] 开始于1， atomicInc 从不实际增加变量 count [0] 超过1。

如果您希望看到它的增量超过1，请为 n 传递较大的值。

变量 n 实际上作为递增过程的滚动值。当要增加的变量实际上达到 n 的值时，下一个 atomicInc 会将其重置为零。 p>

虽然你没有问这个问题，你可能会问：为什么我从来没有看到一个零值，如果我击中了翻转值？

要回答这个问题，你必须记住所有的4个线程都在锁步执行。所有4个执行 atomicInc 指令，然后再执行后续的print语句。

因此， count [0] 从1开始。

3. 第三个线程将其重置为零。


4. 第四个和最后一个线程将其递增为1.

p>

作为另一个实验，尝试启动5个线程，而不是4，看看你是否可以预测打印出来的值。

  ker<<< 1,5>>（hitCount_d）; 
 

由于@talonmies在注释中指出，如果你交换你的 atomicInc for a atomicAdd ：

  int x = atomicAdd （（unsigned int *）& count [0]，n）; 
  

您将得到您可能预期的结果。

I have tried below program using atomicInc().

__global__ void ker(int *count)
{
    int n=1;
    int x = atomicInc ((unsigned int *)&count[0],n);
    CUPRINTF("In kernel count is %d\n",count[0]);
}

int main()
{
    int hitCount[1];
    int *hitCount_d;

    hitCount[0]=1;
    cudaMalloc((void **)&hitCount_d,1*sizeof(int));

    cudaMemcpy(&hitCount_d[0],&hitCount[0],1*sizeof(int),cudaMemcpyHostToDevice);

    ker<<<1,4>>>(hitCount_d);

    cudaMemcpy(&hitCount[0],&hitCount_d[0],1*sizeof(int),cudaMemcpyDeviceToHost);

    printf("count is %d\n",hitCount[0]);
  return 0;
}

Output is:

In kernel count is 1
In kernel count is 1
In kernel count is 1
In kernel count is 1

count is 1

I'm not understanding why it is not incrementing. Can anyone help

解决方案

Referring to the documentation, atomicInc does this:

for the following:

atomicInc ((unsigned int *)&count[0],n);

compute:

((count[0] >= n) ? 0 : (count[0]+1))

and store the result back in count[0]

(If you're not sure what the ? operator does, look here)

Since you've passed n = 1, and count[0] starts out at 1, atomicInc never actually increments the variable count[0] beyond 1.

If you want to see it increment beyond 1, pass a larger value for n.

The variable n actually acts as a "rollover value" for the incrementing process. When the variable to be incremented actually reaches the value of n, the next atomicInc will reset it to zero.

Although you haven't asked the question, you might ask, "Why do I never see a value of zero, if I am hitting the rollover value?"

To answer this, you must remember that all 4 of your threads are executing in lockstep. All 4 of them execute the atomicInc instruction before any execute the subsequent print statement.

Therefore we have a variable of count[0] which starts out at 1.

1. The first thread to execute the atomic resets it to zero.
2. The next thread increments it to 1.
3. The third thread resets it to zero.
4. The fourth and final thread increments it to 1.

Then all 4 threads print out the value.

As another experiment, try launching 5 threads instead of 4, see if you can predict what the value printed out will be.

ker<<<1,5>>>(hitCount_d);

As @talonmies indicated in the comments, if you swap your atomicInc for an atomicAdd:

int x = atomicAdd ((unsigned int *)&count[0],n);

You'll get results that you were probably expecting.

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