Question

我想在Julia中使用共享内存多线程。正如Threads。@ threads宏所做的那样，我可以使用ccall（：jl_threading_run ...）来执行此操作。尽管我的代码现在可以并行运行，但我没有达到我期望的加速。

以下代码旨在作为我所采用的方法和所遇到的性能问题的一个最小示例：[编辑：稍后请参见更多最小示例]

nthreads = Threads.nthreads()
test_size = 1000000
println("STARTED with ", nthreads, " thread(s) and test size of ", test_size, ".")
# Something to be processed:
objects = rand(test_size)
# Somewhere for our results
results = zeros(nthreads)
counts = zeros(nthreads)
# A function to do some work.
function worker_fn()
    work_idx = 1
    my_result = results[Threads.threadid()]
    while work_idx > 0
        my_result += objects[work_idx]
        work_idx += nthreads
        if work_idx > test_size
            break
        end
        counts[Threads.threadid()] += 1
    end
end

# Call our worker function using jl_threading_run
@time ccall(:jl_threading_run, Ref{Cvoid}, (Any,), worker_fn)

# Verify that we made as many calls as we think we did.
println("\nCOUNTS:")
println("\tPer thread:\t", counts)
println("\tSum:\t\t", sum(counts))

在i7-7700上，典型的单线程结果是：

STARTED with 1 thread(s) and test size of 1000000.
 0.134606 seconds (5.00 M allocations: 76.563 MiB, 1.79% gc time)

COUNTS:
    Per thread:     [999999.0]
    Sum:            999999.0

有4个线程：

STARTED with 4 thread(s) and test size of 1000000.
  0.140378 seconds (1.81 M allocations: 25.661 MiB)

COUNTS:
    Per thread:     [249999.0, 249999.0, 249999.0, 249999.0]
    Sum:            999996.0

多线程会使事情变慢！为什么？

编辑：可以在@threads宏本身中创建一个更好的最小示例。

a = zeros(Threads.nthreads())
b = rand(test_size)
calls = zeros(Threads.nthreads())
@time Threads.@threads for i = 1 : test_size
    a[Threads.threadid()] += b[i]
    calls[Threads.threadid()] += 1
end

我错误地认为@threads宏包含在Julia中将意味着有好处。

Answer 1

您遇到的问题很可能是false sharing。

您可以通过如下方式解决您的问题：将您写的区域分开得足够远（我更改了您的代码，使它们运行的时间更长-我也做了“快速而肮脏的”实现以显示更改的本质）：

function f()
    test_size = 10^8
    a = zeros(Threads.nthreads()*100)
    b = rand(test_size)
    calls = zeros(Threads.nthreads()*100)
    Threads.@threads for i = 1 : test_size
        for j in 1:10
            a[Threads.threadid()*100] += b[i]
            calls[Threads.threadid()*100] += 1
        end
    end
    a, calls
end

或在这样的局部变量上进行每线程累加（这是首选方法，因为它应该一致地更快）：

function getrange(n)
    tid = Threads.threadid()
    nt = Threads.nthreads()
    d , r = divrem(n, nt)
    from = (tid - 1) * d + min(r, tid - 1) + 1
    to = from + d - 1 + (tid ≤ r ? 1 : 0)
    from:to
end

function f()
    test_size = 10^8
    a = zeros(Threads.nthreads())
    b = rand(test_size)
    calls = zeros(Threads.nthreads())
    Threads.@threads for k = 1 : Threads.nthreads()
        local_a = 0.0
        local_c = 0.0
        for i in getrange(test_size)
            for j in 1:10
                local_a += b[i]
                local_c += 1
            end
        end
        a[Threads.threadid()] = local_a
        calls[Threads.threadid()] = local_c
    end
    a, calls
end

还请注意，您可能在具有2个物理核心（只有4个虚拟核心）的计算机上使用4个踏步，因此线程化的收益将不是线性的。

朱莉娅（Julia）：共享内存多线程为什么没有给我提速？

1 个答案: