On Sun, May 23, 2021 at 11:09 AM tapi...@gmail.com <tapir....@gmail.com>
wrote:
> It also escapes if `n` is a local variable.
>
> From the code, it looks, if the capacity of the maked slice is larger than
> 1<<12, then the slice is allocated on heap.
>
Seems possible.
Is there a possibility that, in the "make" implementation, different
> routines are chosen by different capacity arguments?
>
It's possible (but I don't think so), but not insofar as it would impact
whether things end up on the stack or the heap.
>
> On Sunday, May 23, 2021 at 5:03:02 AM UTC-4 axel.wa...@googlemail.com
> wrote:
>
>> Hi,
>>
>> there is no such thing as "possibly escaping". The compiler needs to
>> decide whether to emit the code to reserve stack space for a variable or
>> whether to emit the code to allocate heap-space. That's a binary choice, it
>> will always do one or the other.
>>
>> So, yes, `make([]T, n)` in this example always escapes. I assume the
>> heuristic marks it as escaping because `n` is a non-local variable, so it
>> doesn't prove that `n` is effectively constant. It might even just mark
>> every `make` with a `var` argument as escaping.
>>
>> On Sun, May 23, 2021 at 10:51 AM tapi...@gmail.com <tapi...@gmail.com>
>> wrote:
>>
>>> In the following code, "make([]T, n)" is reported as escaped.
>>> But does it really always escape at run time?
>>> Does the report just mean "make([]T, n) possibly escapes to heap"?
>>>
>>> package main
>>>
>>> type T int
>>>
>>> const K = 1<<13
>>> const N = 1<<12
>>> var n = N
>>> var i = n-1
>>>
>>> func main() {
>>> var r = make([]T, N) // make([]T, N) does not escape
>>> println(r[i])
>>>
>>> var r2 = make([]T, n) // make([]T, n) escapes to heap
>>> println(r2[i])
>>>
>>> var r3 = make([]T, K) // make([]T, K) escapes to heap
>>> println(r3[i])
>>> }
>>>
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