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golang面试题:怎么避免内存逃逸?

小白debug大约 4 分钟

问题

怎么避免内存逃逸

怎么答

runtime/stubs.go:133有个函数叫noescapenoescape可以在逃逸分析中隐藏一个指针。让这个指针在逃逸分析中不会被检测为逃逸

 // noescape hides a pointer from escape analysis.  noescape is
 // the identity function but escape analysis doesn't think the
 // output depends on the input.  noescape is inlined and currently
 // compiles down to zero instructions.
 // USE CAREFULLY!
 //go:nosplit
 func noescape(p unsafe.Pointer) unsafe.Pointer {
     x := uintptr(p)
     return unsafe.Pointer(x ^ 0)
}

举例

  • 通过一个例子加深理解,接下来尝试下怎么通过 go build -gcflags=-m 查看逃逸的情况。
package main

import (
	"unsafe"
)
type A struct {
	S *string
}
func (f *A) String() string {
	return *f.S
}
type ATrick struct {
	S unsafe.Pointer
}
func (f *ATrick) String() string {
	return *(*string)(f.S)
}
func NewA(s string) A {
	return A{S: &s}
}
func NewATrick(s string) ATrick {
	return ATrick{S: noescape(unsafe.Pointer(&s))}
}
func noescape(p unsafe.Pointer) unsafe.Pointer {
	x := uintptr(p)
	return unsafe.Pointer(x ^ 0)
}
func main() {
	s := "hello"
	f1 := NewA(s)
	f2 := NewATrick(s)
	s1 := f1.String()
	s2 := f2.String()
	_ = s1 + s2
}

执行go build -gcflags=-m main.go

$go build -gcflags=-m main.go
# command-line-arguments
./main.go:11:6: can inline (*A).String
./main.go:19:6: can inline (*ATrick).String
./main.go:23:6: can inline NewA
./main.go:31:6: can inline noescape
./main.go:27:6: can inline NewATrick
./main.go:28:29: inlining call to noescape
./main.go:36:6: can inline main
./main.go:38:14: inlining call to NewA
./main.go:39:19: inlining call to NewATrick
./main.go:39:19: inlining call to noescape
./main.go:40:17: inlining call to (*A).String
./main.go:41:17: inlining call to (*ATrick).String
/var/folders/45/qx9lfw2s2zzgvhzg3mtzkwzc0000gn/T/go-build763863171/b001/_gomod_.go:6:6: can inline init.0
./main.go:11:7: leaking param: f to result ~r0 level=2
./main.go:19:7: leaking param: f to result ~r0 level=2
./main.go:24:16: &s escapes to heap
./main.go:23:13: moved to heap: s
./main.go:27:18: NewATrick s does not escape
./main.go:28:45: NewATrick &s does not escape
./main.go:31:15: noescape p does not escape
./main.go:38:14: main &s does not escape
./main.go:39:19: main &s does not escape
./main.go:40:10: main f1 does not escape
./main.go:41:10: main f2 does not escape
./main.go:42:9: main s1 + s2 does not escape

其中主要看中间一小段

./main.go:24:16: &s escapes to heap    //这个是NewA中的,逃逸了
./main.go:23:13: moved to heap: s
./main.go:27:18: NewATrick s does not escape // NewATrick里的s的却没逃逸
./main.go:28:45: NewATrick &s does not escape

解释

  • 上段代码对AATrick同样的功能有两种实现:他们包含一个 string ,然后用 String() 方法返回这个字符串。但是从逃逸分析看ATrick 版本没有逃逸。

  • noescape() 函数的作用是遮蔽输入和输出的依赖关系。使编译器不认为 p 会通过 x 逃逸, 因为 uintptr() 产生的引用是编译器无法理解的。

  • 内置的 uintptr 类型是一个真正的指针类型,但是在编译器层面,它只是一个存储一个 指针地址int 类型。代码的最后一行返回 unsafe.Pointer 也是一个 int

  • noescape()runtime 包中使用 unsafe.Pointer 的地方被大量使用。如果作者清楚被 unsafe.Pointer 引用的数据肯定不会被逃逸,但编译器却不知道的情况下,这是很有用的。

  • 面试中秀一秀是可以的,如果在实际项目中如果使用这种 unsafe 包大概率会被同事打死。不建议使用! 毕竟包的名字就叫做 unsafe, 而且源码中的注释也写明了 USE CAREFULLY!

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