Why does a pointer to a local variable escape to the heap?

问题

Here is my go environment:

[lorneli@localhost GoTest]$ go version
go version go1.9 linux/amd64

Here is my program:

package main

type request struct {
    ID string
    size uint32
    off uint64
}

func main() {
    r := request{}
    iter := interface{}(&r) // &r escapes to heap
    iters := make([]interface{}, 0)
    iters = append(iters, iter)
}

I allocate a request instance and convert the pointer of it to interface{}. But when analysing with flag -gcflags "-m" , I found out the instance escapes to heap when converting. Why this happens?

Here is analysing result:

[lorneli@localhost GoTest]$ go build -gcflags "-m"
# _/mnt/hgfs/vmfolder/workspace/GoTest
./main.go:9:6: can inline main
./main.go:11:21: (interface {})(&r) escapes to heap
./main.go:11:22: &r escapes to heap
./main.go:10:15: moved to heap: r
./main.go:12:15: main make([]interface {}, 0) does not escape

I think this case doesn't match any cases listed on "Go Escape Analysis Flaws".

回答1:

Simplify your example. Analyze with -gcflags='-m -m'.

Example 1:

package main

func main() {
    var v int
    s := make([]*int, 0)
    s = append(s, &v) // &v escapes to heap
}

Output:

$ go version
go version devel +df8c2b905b Tue Mar 6 06:13:17 2018 +0000 linux/amd64
$ go run -gcflags='-m -m' esc.go
# command-line-arguments
./esc.go:3:6: can inline main as: func() { var v int; v = <N>; s := make([]*int, 0); s = append(s, &v) }
./esc.go:6:16: &v escapes to heap
./esc.go:6:16:  from append(s, &v) (appended to slice) at ./esc.go:6:12
./esc.go:4:6: moved to heap: v
./esc.go:5:11: main make([]*int, 0) does not escape
$

Escape analysis determines whether any references to a value escape the function in which the value is declared. A reference to the variable v, declared in function main, escapes as an argument to function append: &v escapes to heap from append(s, &v), moved to heap: v.

Example 2:

package main

func main() {
    var v int
    lc := 1
    s := make([]*int, lc)
    s[0] = &v
}

$ go run -gcflags='-m -m' esc2.go
./esc2.go:3:6: can inline main as: func() { var v int; v = <N>; lc := 1; s := make([]*int, lc); s[0] = &v }
./esc2.go:6:11: make([]*int, lc) escapes to heap
./esc2.go:6:11:     from make([]*int, lc) (too large for stack) at ./esc2.go:6:11
./esc2.go:7:9: &v escapes to heap
./esc2.go:7:9:  from s[0] (slice-element-equals) at ./esc2.go:7:7
./esc2.go:4:6: moved to heap: v
$

type slice struct {
    array unsafe.Pointer
    len   int
    cap   int
}

make for a slice returns a slice descriptor struct (pointer to underlying array, length, and capacity) and allocates an underlying slice element array. The underlying array is generally allocated on the heap: make([]*int, lc) escapes to heap from make([]*int, lc).

s[0] = &v stores a reference to the variable v (&v) in the underlying array on the heap: &v escapes to heap from s[0] (slice-element-equals), moved to heap: v. The reference remains on the heap, after the function ends and its stack is reclaimed, until the underlying array is garbage collected.

If the make slice capacity is a small (compile time) constant, make([]*int, 1) in your example, the underlying array may be allocated on the stack. However, escape analysis does not take this into account.

来源：https://stackoverflow.com/questions/49125779/why-does-a-pointer-to-a-local-variable-escape-to-the-heap

标签

heap