On Sunday, May 23, 2021 at 5:42:41 AM UTC-4 tapi...@gmail.com wrote:
> On Sunday, May 23, 2021 at 4:51:30 AM UTC-4 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])
>> }
>>
>
> Another question is, why should "make([]T, K)" escape to heap?
> Using the capacity as the criterion is not reasonable.
> After all arrays with even larger sizes will not allocated on heap.
>
It looks, capacity is not only criterion. Whether or not there is a pointer
referencing the allocated elements is another factor.
In the following program, if K is changed to "1<<12", then no escapes.
But, this is still not reasonable.
package main
const K = 1<<13
var i = K-1
func main() {
var x = new([K]int) // new([8192]int) escapes to heap
println(x[i])
var y [K]int // not escape
println(y[i])
var z = [K]int{} // not escape
println(z[i])
var w = &[K]int{} // &[8192]int{} escapes to heap
println(w[i])
}
>
>
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