Multithreading 使用Julia绘制带有多线程的mandelbrot，竞争条件问题

我是Julia的新手，我正在尝试实现Julia的多线程，但我相信我遇到了“竞争条件问题”。这里我正在绘制mandelbrot图，但我相信由于竞争条件，数组索引[n]会干扰颜色映射。我尝试将原子特性用于索引n，但显然我不能将该类型用作索引。下面是要比较的图片以及代码块

谢谢

module MandelBrot
using Plots
#make some functions for mandelbrot stuff
#find out if a number is part of the set
#remember the mandelbrot is symmetrical about the real number plane

function mandel(c)
    #determine if a number is in the set or not in the set - 
    max_iter = 1000;
    bound = 2
    z = 0
    n = 0
    #if the magnitude of z exceeds two we know we are done.
    while abs(z)<bound && n<max_iter
        z = z^2+c
        n+=1
    end
    return n #if n is 1000 assume c is good, else not of the set
end

#map n to a color
function brot(n)
    rgb = 250
    m = (n%rgb) /rgb#divide 250
    if 0< n <= 250
        c = RGB(1,m,0)
    elseif 250<n<=500
        c = RGB(1-m,1,0)
    elseif 500<n<=750
        c = RGB(0,1,m)
    elseif 750<n<=999
        c = RGB(0,1-m,1)
    else
        c=RGB(0,0,0)
    end
    return c
    #TODO: append this c to an array of color values
end
#mrandom


function mandelbrot(reals,imags)
    #generate #real amount of points between -2 and 1
    #and #imag amount of points between 0 and i
    #determine if any of those combinations are in the mandelbrot set
    r = LinRange(-2,1,reals)
    i = LinRange(-1,1,imags)
    master_list = zeros(Complex{Float64},reals*imags,1)
    color_assign = Array{RGB{Float64}}(undef,reals*imags,1)
    #n = Threads.Atomic{Int64}(1)
    n = 1
        Threads.@threads for real_num in r
            for imaginary_num in i
                #z = complex(real_num, imaginary_num) #create the number
                #master_list[n] = z                      #add it to the list
                #color_assign[n,1] = (brot ∘ mandel)(z) #function of function! \circ + tab

                #or would this be faster? since we dont change z all the time?
                master_list[n] = complex(real_num, imaginary_num)
                color_assign[n,1] = (brot ∘ mandel)(complex(real_num, imaginary_num))
                n+=1
                #Threads.atomic_add!(n,1)
            end
        end
        gr(markerstrokewidth=0,markerstrokealpha=0,markersize=.5,legend=false)
        scatter(master_list,markerstrokecolor=color_assign,color=color_assign,aspect_ratio=:equal)
    end

#end statement for the module
end

julia> @time m.mandelbrot(1000,1000)
  2.260481 seconds (6.01 M allocations: 477.081 MiB, 9.56% gc time)

模块MandelBrot
使用绘图
#为mandelbrot的东西做一些函数
#找出一个数字是否是集合的一部分
#记住mandelbrot关于实数平面是对称的
曼德尔函数（c）
#确定某个数字是否在集合中-
最大电阻=1000；
界限=2
z=0
n=0
#如果z的大小超过2，我们就知道我们完成了。
当abs（z）时，以下是应该有帮助的：

function mandelbrot(reals,imags)
    r = LinRange(-2,1,reals)
    i = LinRange(0,1,imags)
    master_list = zeros(Complex{Float64},reals*imags,1)
    color_assign = Array{RGB{Float64}}(undef,reals*imags,1)
    Threads.@threads for a in 1:reals
        real_num = r[a]
        for (b, imaginary_num) in enumerate(i)
            n = (a-1)*imags + b
            master_list[n] = complex(real_num, imaginary_num)
            color_assign[n, 1] = (brot ∘ mandel)(complex(real_num, imaginary_num))
        end
    end
    gr(markerstrokewidth=0,markerstrokealpha=0,markersize=1,legend=false)
    scatter(master_list,markerstrokecolor=color_assign,color=color_assign,aspect_ratio=:equal)
end

该方法是将
n
计算为
r
和
i
沿线指数的函数。
还要注意，我使用
1:reals
而不仅仅是
枚举（r）
作为
线程。@Threads
不接受任意迭代器

请注意，您的代码可能会在其他版本中进行清理，但如果没有完全可复制的示例，则很难做到这一点。
谢谢您，卡明斯基先生！我将很快尝试一下-我添加了完整的代码以保证可复制性，很抱歉之前没有这么做。我刚刚检查过它-所有这些都可以在多个线程上运行。只需将其命名为
MandelBrot.MandelBrot（10001000）
。