2013-08-07 86 views
0

我正在尝试编写一些CUDA代码来计算最长的公用子序列。我不能工作了如何使线程休眠,直到依存关系来计算它的细胞被满足:睡觉/在CUDA线程中等待

// 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; 
} 

所以基本上线程应该在while循环挂起,直到它的依赖,满足在移入计算代码之前由其他线程执行。

我知道“第一”线程都有它的依赖性来满足它打印

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 

曾经它已经完成它的计算设置了一些细胞依赖矩阵为true,允许2个线程,让过去的同时,循环和内核从那里移动。

但是,如果我去掉while循环我的整个系统挂起〜10秒,我得到

the launch timed out and was terminated 

有什么建议?

回答

1

睡眠不好主意,更好地等待条件变量或互斥锁。

在GPU上,每个条件语句都非常昂贵。所以如果可以的话,尝试并行化所有代码。为了确保代码被完成了所有的线程可以使用__syncthreads()

如果你还是想用最简单的方法添加互斥体,但它通常坏主意