问题
怎么避免内存逃逸?
怎么答
在runtime/stubs.go:133
有个函数叫noescape
。noescape
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 mainimport ( "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
A
和ATrick
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
的int
类型。代码的最后一行返回unsafe.Pointer
也是一个int
。noescape()
在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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文章来源:智云一二三科技
文章标题:golang面试题:怎么避免内存逃逸?
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