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.
Is there a possibility that, in the "make" implementation, different
routines are chosen by different capacity arguments?
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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