On Sun, Jan 27, 2019 at 6:21 PM wrote:
>
> It is in line 899 and line 914 of the malloc.go file.
>
> 899 spc := makeSpanClass(sizeclass, noscan)
> 914 s = largeAlloc(size, needzero, noscan)
Thanks, I wasn't aware of that. Looks like we have spans that contain
only objects without pointers, and
It is in line 899 and line 914 of the malloc.go file.
899 spc := makeSpanClass(sizeclass, noscan)
914 s = largeAlloc(size, needzero, noscan)
在 2019年1月27日星期日 UTC+8下午1:55:29，Ian Lance Taylor写道：
>
> On Sat, Jan 26, 2019 at 6:50 PM > wrote:
> >
> > I am looking at go1.11.1.
>
> I don't see a
On Sat, Jan 26, 2019 at 6:50 PM wrote:
>
> I am looking at go1.11.1.
I don't see a "noscan" flag for mallocgc in 1.11.1. There is a noscan
local variable, which is set of the type is passed as nil (the case I
mentioned earlier) or if the type has no pointers.
> The memory allocation is still
I am looking at go1.11.1.
The memory allocation is still difficult to understand. What reference
materials can help me understand the context of this block?
在 2019年1月27日星期日 UTC+8上午3:03:47，Ian Lance Taylor写道：
>
> On Sat, Jan 26, 2019 at 5:37 AM > wrote:
> >
> > Why the garbage collector won't
On Sat, Jan 26, 2019 at 5:37 AM wrote:
>
> Why the garbage collector won't know how to find the pointers?
> I looked at mallocgc and decided if the GC needs to scan this object based
> on the noscan flag.
What version of Go are you looking at? There was a flagNoScan as late
as Go 1.6, but
Why the garbage collector won't know how to find the pointers?
I looked at mallocgc and decided if the GC needs to scan this object based
on the noscan flag.
在 2019年1月25日星期五 UTC+8下午10:58:44，Ian Lance Taylor写道：
>
> On Thu, Jan 24, 2019 at 11:58 PM >
> wrote:
> >
> > go 1.11.1 source code is
On Thu, Jan 24, 2019 at 11:58 PM wrote:
>
> go 1.11.1 source code is below:
>
> Generally speaking, make chan just pay attention to the presence or absence
> of buf.
>
> When I saw the source code of make chan, I can understand case 1: chan buf
> is 0, but can't understand case 2 & default.
>
go 1.11.1 source code is below:
Generally speaking, make chan just pay attention to the presence or absence
of buf.
When I saw the source code of make chan, I can understand case 1: chan buf
is 0, but can't understand case 2 & default.
Who knows this principle?
Thanks!
var c *hchan
