julia - 别名元组类型时的额外内存分配

Question

使用 Julia 1.5.3，我有以下代码(当然它不是实际代码，只是一个人为的示例，我可以用它来重现问题):

function basic()
    result::Tuple{Int64,Int64,Int64} = (0, 0, 0)
    for i in 1:100000
        for i in -1:1, j in -1:1, k in -1:1
            result = result .+ (i, j, k)
        end
    end
    result
end

function fancy(T)
    result::T = (0, 0, 0)
    for i in 1:100000
        for i in -1:1, j in -1:1, k in -1:1
            result = result .+ (i, j, k)::T
        end
    end
    result
end

# Warmup
for i in 1:10
    basic()
    fancy(Tuple{Int64,Int64,Int64})
end

println("basic:")
println(@time basic())
println("fancy:")
println(@time fancy(Tuple{Int64,Int64,Int64}))

输出:

basic:
  0.000000 seconds
(0, 0, 0)
fancy:
  0.762599 seconds (10.80 M allocations: 494.385 MiB, 3.68% gc time)
(0, 0, 0)

如果我改写:

            result = result .+ (i, j, k)::T

我得到:

basic:
  0.000000 seconds
(0, 0, 0)
fancy:
  0.789512 seconds (13.50 M allocations: 576.782 MiB, 3.77% gc time)
(0, 0, 0)

有没有办法让 basic和 fancy执行相同的操作，即在 fancy 中没有分配?

Answer 1

只需更改 fancy 的签名即可:

function fancy(::Type{T}) where T
    result::T = (0, 0, 0)
    for i in 1:100000
        for i in -1:1, j in -1:1, k in -1:1
            result = result .+ (i, j, k)::T
        end
    end
    result
end

println(@time fancy(Tuple{Int64,Int64,Int64}))

请注意，这两个函数实际上都优化为几乎无操作。您可以通过 @code_native basic() 检查它.