Sorry it seems that I simplified my problem too much in the previous code.
Actually I am running something similar to the following, and I run it as a
function and it starts to have allocation since i=512. I am really unsure
if it's my coding problem(so it can be improved to 0 bytes) or it is
natural in the system that allocations appear. Thank you so much!
function test()
B=fill!(cell(5,5),Int64[])
for i=1:513
push!(B[2,3],i)
end
T=zeros(Int64,513)
for i=1:513
@time T[i]=B[2,3][i]
println(T[i])
end
end
在 2014年12月1日星期一UTC-6下午7时24分31秒，John Myles White写道：
>
> Did you run this inside a function? If not, your results are not going to
> be useful indicators of how code will perform inside a function.
>
> Inside of a function body, I see 0 bytes being allocated.
>
> -- John
>
> On Dec 1, 2014, at 4:53 PM, Yijing Wu <[email protected] <javascript:>>
> wrote:
>
> Hi all, I found a strange problem about a very simple code in julia and
> hopefully I can get some help from you, here is the code:
>
>
> B=[1:1000]
>
> T=zeros(Int64,1000)
>
> for i=1:1000
>
> @time T[i]=B[i]
>
> println(T[i])
>
> end
>
>
> And when I run the code, the @time shows that it require 48 bytes allocation
> when i is larger than or equal to 512, and 0 bytes when smaller. Is this a
> problem that can be improved or I have to accept that it is designed to take
> some allocations when larger than 512?
>
>
> Thanks a lot for your help!
>
>
>
