为什么比较结构数组有不同的结果 [英] why struct arrays comparing has different result
问题描述
我不知道为什么会这样,而且我找不到相对的源代码.有人可以向我解释吗?
I don't know why the below happens, and I can't find source code relative. Can anybody explain to me?
var s, ss struct{} // two empty structs
arr1 := [6]*struct{}{&s} // array with empty struct pointer
arr2 := [6]*struct{}{&ss} // array with empty struct pointer
fmt.Println(&s == &ss, arr1 == arr2) // false, true
var l, ll struct{A int}{}
arr3 := [6]*struct{A int}{&l} // array with empty struct pointer
arr4 := [6]*struct{A int}{&ll} // array with empty struct pointer
fmt.Println(&l == &ll, arr3 == arr4) // false, false
推荐答案
指针值具有可比性.如果两个指针值指向同一个变量,或者两个指针值都为
nil
,则它们相等.指向不同的零大小变量的指针可能相等,也可能不相等.
Pointer values are comparable. Two pointer values are equal if they point to the same variable or if both have value
nil
. Pointers to distinct zero-size variables may or may not be equal.
如果结构或数组类型不包含大小大于零的字段(或分别包含元素),则其大小为零.两个不同的零大小变量可能在内存中具有相同的地址.
A struct or array type has size zero if it contains no fields (or elements, respectively) that have a size greater than zero. Two distinct zero-size variables may have the same address in memory.
s
和 ss
变量的大小为零,因此& s
和&ss
为指向零大小变量的指针,因此规范不保证它们的相等性.这意味着& s ==& ss
可能会评估为 true
或 false
,您不能指望结果将会是,这样做是错误的.
Size of the s
and ss
variables are zero, so &s
and &ss
are pointers to distinct zero-size variables, so the spec does not guarantee anything about their equality. What this means is that &s == &ss
may evaluate to either true
or false
, you can't count on what the result will be, and it would be a mistake to do so.
仍然,很奇怪的是,在应用程序的单个运行时,一旦它们相等,一次就不相等.教训是永远不要依赖它.
Still, it is weird that during a single runtime of the app, once they are equal, and once they are not. Lesson is to not rely on it, ever.
通过逃逸分析可以解释不同的行为.
The different behavior can be explained by looking at the escape analysis.
让我们将您的应用程序简化为:
Let's simplify your app to this:
var s, ss struct{} // two empty structs
arr1 := [6]*struct{}{&s} // array with empty struct pointer
arr2 := [6]*struct{}{&ss} // array with empty struct pointer
fmt.Println(&s == &ss, arr1 == arr2) // false, true
使用 go运行转义分析-gcflags'-m'play.go
给出:
./play.go:13:17: &s == &ss escapes to heap
./play.go:13:30: arr1 == arr2 escapes to heap
./play.go:11:23: main &s does not escape
./play.go:12:23: main &ss does not escape
./play.go:13:14: main &s does not escape
./play.go:13:20: main &ss does not escape
./play.go:13:13: main ... argument does not escape
false true
& s
和& ss
不会转义(因为它们不会传递给 fmt.Println()
,只会传递结果的& s ==& ss
).
&s
and &ss
do not escape (as they are not passed to fmt.Println()
, only the result of &s == &ss
).
如果我们在上述简化的应用中添加一行:
If we add a single line to the above simplified app:
var s, ss struct{} // two empty structs
arr1 := [6]*struct{}{&s} // array with empty struct pointer
arr2 := [6]*struct{}{&ss} // array with empty struct pointer
fmt.Println(&s == &ss, arr1 == arr2) // true, true
fmt.Printf("%p %p\n", &s, &ss) // true, true
运行逃逸分析现在可以得出:
Running escape analysis now gives:
./play.go:13:17: &s == &ss escapes to heap
./play.go:13:30: arr1 == arr2 escapes to heap
./play.go:15:24: &s escapes to heap
./play.go:15:24: &s escapes to heap
./play.go:10:6: moved to heap: s
./play.go:15:28: &ss escapes to heap
./play.go:15:28: &ss escapes to heap
./play.go:10:9: moved to heap: ss
./play.go:11:23: main &s does not escape
./play.go:12:23: main &ss does not escape
./play.go:13:14: main &s does not escape
./play.go:13:20: main &ss does not escape
./play.go:13:13: main ... argument does not escape
./play.go:15:12: main ... argument does not escape
true true
行为发生了变化:我们现在看到 true true
输出(在上尝试一下)进入游乐场).
The behavior changed: we now see true true
output (try it on the Go Playground).
更改行为的原因是因为& s
和& ss
逃逸到堆中:它们直接传递给 fmt.Println()
,因此编译器更改了它们的存储方式(位置),并更改了它们的地址.
The reason for the changed behavior is because &s
and &ss
escape to heap: they are directly passed to fmt.Println()
, so the compiler changed how (where) they are stored, and with that, so did their address.
查看相关的/可能的重复项:空结构
See related / possible duplicate: Golang Address of Slices of Empty Structs
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