在Go语言中，关于堆栈分配，什么被认为是“小”对象？

huangapple
go评论114阅读模式
英文:

What is considered "small" object in Go regarding stack allocation?

问题

代码中的a和b都是切片，但它们的长度不同。根据Go语言的规则，长度小于等于_MaxSmallSize的切片会在栈上分配内存，而长度大于_MaxSmallSize的切片会在堆上分配内存。

在你的代码中，a的长度是8191，小于_MaxSmallSize，所以它被分配在栈上。而b的长度是8192，大于_MaxSmallSize，所以它被分配在堆上。

_MaxSmallSize的值是32 << 10，即32乘以1024，等于32768，也就是32KB。因此，长度小于等于32768的切片会在栈上分配内存，长度大于32768的切片会在堆上分配内存。

关于make函数的使用，当分配的内存大小大于64KB时，切片会在堆上分配内存，否则会在栈上分配内存。

以上是关于为什么a是小对象而b是大对象的解释。

英文:

The code:

  1. func MaxSmallSize() {
  2. 	a := make([]int64, 8191)
  3. 	b := make([]int64, 8192)
  4. 	_ = a
  5. 	_ = b
  6. }

Then run go build -gcflags=&#39;-m&#39; . 2&gt;&amp;1 to check memory allocation details. The result:

  1. ./mem.go:10: can inline MaxSmallSize
  2. ./mem.go:12: make([]int64, 8192) escapes to heap
  3. ./mem.go:11: MaxSmallSize make([]int64, 8191) does not escape

My question is why a is small object and b is large object?

make 64KB will escape to heap and less will allocate in stack. Does the _MaxSmallSize = 32 &lt;&lt; 10 is the reason?

go env

  1. GOARCH=&quot;amd64&quot;
  2. GOBIN=&quot;&quot;
  3. GOEXE=&quot;&quot;
  4. GOHOSTARCH=&quot;amd64&quot;
  5. GOHOSTOS=&quot;linux&quot;
  6. GOOS=&quot;linux&quot;
  7. GOPATH=&quot;/vagrant/gopath&quot;
  8. GORACE=&quot;&quot;
  9. GOROOT=&quot;/home/vagrant/go&quot;
  10. GOTOOLDIR=&quot;/home/vagrant/go/pkg/tool/linux_amd64&quot;
  11. CC=&quot;gcc&quot;
  12. GOGCCFLAGS=&quot;-fPIC -m64 -pthread -fmessage-length=0 -fdebug-prefix-map=/tmp/go-build201775001=/tmp/go-build&quot;
  13. CXX=&quot;g++&quot;
  14. CGO_ENABLED=&quot;1&quot;

答案1

得分: 3

由于这在语言规范中没有提到，它是一个实现细节，因此可能会根据多种因素（Go版本、目标操作系统、架构等）而有所不同。

如果你想找出它的当前值或者开始查找的地方，请查看cmd/compile/internal/gc包。

决定变量分配位置的逃逸分析在cmd/compile/internal/gc/esc.go中。检查切片的make操作在未导出的函数esc()中：

  1. func esc(e *EscState, n *Node, up *Node) {
  2.     // ...
  4.     // Big stuff escapes unconditionally
  5.     // "Big" conditions that were scattered around in walk have been gathered here
  6.     if n.Esc != EscHeap && n.Type != nil &&
  7.         (n.Type.Width > MaxStackVarSize ||
  8.             (n.Op == ONEW || n.Op == OPTRLIT) && n.Type.Elem().Width >= 1<<16 ||
  9.             n.Op == OMAKESLICE && !isSmallMakeSlice(n)) {
  10.         if Debug['m'] > 2 {
  11.             Warnl(n.Lineno, "%v is too large for stack", n)
  12.         }
  13.         n.Esc = EscHeap
  14.         addrescapes(n)
  15.         escassignSinkNilWhy(e, n, n, "too large for stack") // TODO category: tooLarge
  16.     }
  18.     // ...
  19. }

涉及大小的决策在函数isSmallMakeSlice()中，它位于文件cmd/compile/internal/gc/walk.go中：

  1. func isSmallMakeSlice(n *Node) bool {
  2.     if n.Op != OMAKESLICE {
  3.         return false
  4.     }
  5.     l := n.Left
  6.     r := n.Right
  7.     if r == nil {
  8.         r = l
  9.     }
  10.     t := n.Type
  12.     return Smallintconst(l) && Smallintconst(r) && (t.Elem().Width == 0 || r.Int64() < (1<<16)/t.Elem().Width)
  13. }

大小限制如下：

  1. r.Int64() < (1<<16)/t.Elem().Width

r是切片的长度或容量（如果提供了容量），t.Elem().Width是元素类型的字节大小：

  1. NumElem < 65536 / SizeElem

在你的情况下：

  1. NumElem < 65536 / 8 = 8192

因此，如果切片类型是[]uint64，则从8192开始，它将在堆上分配（而不是栈上），就像你遇到的情况一样。

英文:

Since this is not mentioned in the language spec, it is an implementation detail, and as such, it may vary based on a number of things (Go version, target OS, architecture etc.).

If you want to find out its current value or a place to start digging, check out the cmd/compile/internal/gc package.

The escape analysis which decides where to allocate the variable is in cmd/compile/internal/gc/esc.go. Check of the make slice operation is in unexported function esc():

  1. func esc(e *EscState, n *Node, up *Node) {
  2. 	// ...
  4. 	// Big stuff escapes unconditionally
  5. 	// &quot;Big&quot; conditions that were scattered around in walk have been gathered here
  6. 	if n.Esc != EscHeap &amp;&amp; n.Type != nil &amp;&amp;
  7. 		(n.Type.Width &gt; MaxStackVarSize ||
  8. 			(n.Op == ONEW || n.Op == OPTRLIT) &amp;&amp; n.Type.Elem().Width &gt;= 1&lt;&lt;16 ||
  9. 			n.Op == OMAKESLICE &amp;&amp; !isSmallMakeSlice(n)) {
  10. 		if Debug[&#39;m&#39;] &gt; 2 {
  11. 			Warnl(n.Lineno, &quot;%v is too large for stack&quot;, n)
  12. 		}
  13. 		n.Esc = EscHeap
  14. 		addrescapes(n)
  15. 		escassignSinkNilWhy(e, n, n, &quot;too large for stack&quot;) // TODO category: tooLarge
  16. 	}
  18. 	// ...
  19. }

The decision involving the size is in function isSmallMakeSlice(), this is in file cmd/compile/internal/gc/walk.go:

  1. func isSmallMakeSlice(n *Node) bool {
  2. 	if n.Op != OMAKESLICE {
  3. 		return false
  4. 	}
  5. 	l := n.Left
  6. 	r := n.Right
  7. 	if r == nil {
  8. 		r = l
  9. 	}
  10. 	t := n.Type
  12. 	return Smallintconst(l) &amp;&amp; Smallintconst(r) &amp;&amp; (t.Elem().Width == 0 || r.Int64() &lt; (1&lt;&lt;16)/t.Elem().Width)
  13. }

The size limit is this:

  1. r.Int64() &lt; (1&lt;&lt;16)/t.Elem().Width

r is the length or capacity of the slice (if cap is provided), t.Elem().Width is the byte size of the element type:

  1. NumElem &lt; 65536 / SizeElem

In your case:

  1. NumElem &lt; 65536 / 8 = 8192

So if the slice type is []uint64, 8192 is the limit from which it is allocated on the heap (instead of the stack), just as you experienced.

答案2

得分: 2

@icza的回答非常有见地，我只想补充一点，链接有点过时了，5年后你可以在cmd/compile/internal/escape/utils.go和cmd/compile/internal/ir/cfg.go中找到代码：

  1. // HeapAllocReason返回给定节点必须分配在堆上的原因，如果不需要分配则返回空字符串。
  2. func HeapAllocReason(n ir.Node) string {
  3. 	// ... 省略部分代码
  4. 	if n.Op() == ir.OMAKESLICE {
  5. 		n := n.(*ir.MakeExpr)
  6. 		r := n.Cap
  7. 		if r == nil {
  8. 			r = n.Len
  9. 		}
  10. 		if !ir.IsSmallIntConst(r) {
  11. 			return "非常数大小"
  12. 		}
  13. 		if t := n.Type(); t.Elem().Size() != 0 && ir.Int64Val(r) > ir.MaxImplicitStackVarSize/t.Elem().Size() {
  14. 			return "太大，无法放在栈上"
  15. 		}
  16. 	}
  18. 	return ""
  19. }

而ir.MaxImplicitStackVarSize是：

  1. package ir
  3. var (
  4. 	// 在栈上分配的隐式变量的最大大小。
  5. 	//   p := new(T)          在栈上分配T
  6. 	//   p := &T{}            在栈上分配T
  7. 	//   s := make([]T, n)    在栈上分配[n]T
  8. 	//   s := []byte("...")   在栈上分配[n]byte
  9. 	// 注意：标志smallframes可以更新此值。
  10. 	MaxImplicitStackVarSize = int64(64 * 1024)
  11. )
英文:

@icza's answer is really insightful, I'd just like to add that the link is a bit outdated 5 years later, you can find the code in cmd/compile/internal/escape/utils.go and in turn cmd/compile/internal/ir/cfg.go now:

  1. // HeapAllocReason returns the reason the given Node must be heap
  2. // allocated, or the empty string if it doesn&#39;t.
  3. func HeapAllocReason(n ir.Node) string {
  4. 	// ... omitted for brevity
  5. 	if n.Op() == ir.OMAKESLICE {
  6. 		n := n.(*ir.MakeExpr)
  7. 		r := n.Cap
  8. 		if r == nil {
  9. 			r = n.Len
  10. 		}
  11. 		if !ir.IsSmallIntConst(r) {
  12. 			return &quot;non-constant size&quot;
  13. 		}
  14. 		if t := n.Type(); t.Elem().Size() != 0 &amp;&amp; ir.Int64Val(r) &gt; ir.MaxImplicitStackVarSize/t.Elem().Size() {
  15. 			return &quot;too large for stack&quot;
  16. 		}
  17. 	}
  19. 	return &quot;&quot;
  20. }

and ir.MaxImplicitStackVarSize is:

  1. package ir
  3. var (
  4. 	// maximum size of implicit variables that we will allocate on the stack.
  5. 	//   p := new(T)          allocating T on the stack
  6. 	//   p := &amp;T{}            allocating T on the stack
  7. 	//   s := make([]T, n)    allocating [n]T on the stack
  8. 	//   s := []byte(&quot;...&quot;)   allocating [n]byte on the stack
  9. 	// Note: the flag smallframes can update this value.
  10. 	MaxImplicitStackVarSize = int64(64 * 1024)
  11. )

huangapple
  • 本文由 发表于 2017年2月15日 15:34:41
  • 转载请务必保留本文链接：https://go.coder-hub.com/42243197.html
  • allocation
  • escape-analysis
  • go
  • slice
匿名

发表评论

匿名网友

:?: :razz: :sad: :evil: :!: :smile: :oops: :grin: :eek: :shock: :???: :cool: :lol: :mad: :twisted: :roll: :wink: :idea: :arrow: :neutral: :cry: :mrgreen:

确定