Question

我正在尝试编写一些CUDA代码来计算最长的公共子序列。我无法弄清楚如何使线程休眠，直到满足计算它的单元格的依赖关系：

即

// Ignore the spurious maths here, very messy data structures. Planning ahead to strings that are bigger then GPU blocks. i & j are correct though.
int real_i = blockDim.x * blockIdx.x + threadIdx.x;
int real_j = blockDim.y * (max_offset - blockIdx.x) + threadIdx.y;

char i_char = seq1[real_i];
char j_char = seq2[real_j];

// For i & j = 1 to length
if((real_i > 0 && real_j > 0) && (real_i < sequence_length && real_j < sequence_length) {

    printf("i: %d, j: %d\n", real_i, real_j);
    printf("I need to wait for dependancy at i: %d j: %d and i: %d j: %d\n", real_i, (real_j - 1), real_i - 1, real_j);
    printf("Is this true? %d\n", (depend[sequence_length * real_i + (real_j - 1)] && depend[sequence_length * (real_i - 1) + real_j]));

    //WAIT FOR DEPENDENCY TO BE SATISFIED
    //THIS IS WHERE I NEED THE CODE TO HANG
    while( (depend[sequence_length * real_i + (real_j - 1)] == false) && (depend[sequence_length * (real_i - 1) + real_j] == false) ) {
    }

    if (i_char == j_char)
        c[sequence_length * real_i + real_j] = (c[sequence_length * (real_i - 1) + (real_j - 1)]) + 1;
     else
        c[sequence_length * real_i + real_j] = max(c[sequence_length * real_i + (real_j - 1)], c[sequence_length * (real_i - 1) + real_j]);

    // SETTING THESE TO TRUE SHOULD ALLOW OTHER THREADS TO BREAK PAST THE WHILE BLOCK
    depend[sequence_length * real_i + (real_j - 1)] = true;
    depend[sequence_length * (real_i - 1) + real_j] = true;
}

基本上，线程应该挂在while循环上，直到其依赖关系，在进入计算代码之前被其他线程满足。

我知道'first'线程在打印时满足其依赖性

real i 1, real j 1
I need to wait for dependancy at i: 1 j: 0 and i: 0 j: 1
Is this true? 1

一旦完成计算，将依赖矩阵中的一些单元格设置为true，允许另外2个线程通过while循环，内核从那里移动。

但是，如果我取消注释while循环，整个系统会挂起约10秒钟，然后我会

the launch timed out and was terminated

有什么建议吗？

Answer 1

睡觉是个坏主意，最好等待条件变量或互斥量。

在GPU上，每个条件声明都非常昂贵。所以，如果可以的话，尝试并行化所有代码。要确保代码在所有线程中完成，您可以使用__syncthreads()

如果您仍想使用最简单的解决方案添加互斥锁，但这通常是个坏主意

在CUDA线程中睡觉/等待

1 个答案: