Question

我有一个数组需要包含不同内容的总和，因此我想对每个元素执行减少。这是代码：

cdef int *a=<int *>malloc(sizeof(int) * 3)
for i in range(3):
    a[i]=1*i
cdef int *b
for i in prange(1000,nogil=True,num_threads=10):
    b=res() #res returns an array initialized to 1s
    with gil: #if commented this line gives erroneous results 
        for k in range(3):
            a[k]+=b[k]
for i in range(3):
    print a[i]

直到 with gil 代码运行正常，否则会产生错误的结果。如何处理数组的每个元素的减少而不使用gil导致gil我认为将阻止其他线程

Answer 1

减少通常在实践中起作用的方式是为每个线程单独执行求和，然后在最后将它们一起添加。你可以用

之类的东西手动完成

cdef int *b
cdef int *a_local # version of a that is duplicated by each thread
cdef int i,j,k

# set up as before
cdef int *a=<int *>malloc(sizeof(int) * 3)
for i in range(3):
    a[i]=1*i

# multithreaded from here
with nogil, parallel(num_threads=10):
    # setup and initialise a_local on each thread
    a_local = <int*>malloc(sizeof(int)*3)
    for k in range(3):
        a_local[k] = 0

    for i in prange(1000):
        b=res() # Note - you never free b
                # this is likely a memory leak....

        for j in range(3):
            a_local[j]+=b[j]

    # finally at the end add them all together.
    # this needs to be done `with gil:` to avoid race conditions 
    # but it isn't a problem
    # because it's only a small amount of work being done
    with gil:
        for k in range(3):
            a[k] += a_local[k]
    free(a_local)

在cython并行中减少数组

1 个答案: