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 <tapir....@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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