英文:
Are maps passed by value or by reference in Go?
问题
在Go语言中,地图(maps)是通过引用传递的。这意味着当你将地图作为参数传递给函数时,实际上是传递了指向地图的指针,而不是地图的副本。因此,对地图的任何修改都会影响到原始地图。
对于返回值,你可以选择返回地图的指针或者将地图作为值返回。如果你返回地图的指针,那么调用者将可以直接修改地图的内容。如果你返回地图作为值,那么调用者将得到地图的副本,对副本的修改不会影响到原始地图。
关于性能方面的考虑,如果地图较大,且你希望避免不必要的数据复制,那么返回地图的指针可能更高效。但是需要注意的是,返回指针也意味着调用者可能会修改原始地图的内容,这可能会引入潜在的并发问题。
总的来说,选择是根据你的具体需求和代码设计来决定的。如果你不确定,可以根据实际情况进行测试和性能分析,以选择最适合的方式。
英文:
Are maps passed by value or reference in Go ?
It is always possible to define a function as following, but is this an overkill ?
func foo(dat *map[string]interface{}) {...}
Same question for return value. Should I return a pointer to the map, or return the map as value ?
The intention is of course to avoid unnecessary data copy.
答案1
得分: 143
在这个线程中,你会找到你的答案:
https://stackoverflow.com/questions/28384343/golang-accessing-a-map-using-its-reference
> 你不需要在 map 中使用指针。
>
> Map 类型是引用类型,就像指针或切片一样1。
>
> 如果你需要改变 Session,你可以使用指针:
>
> map[string]*Session
>
> https://blog.golang.org/go-maps-in-action
英文:
In this thread you will find your answer :
https://stackoverflow.com/questions/28384343/golang-accessing-a-map-using-its-reference
> You don't need to use a pointer with a map.
>
> Map types are reference types, like pointers or slices1
>
> If you needed to change the Session you could use a pointer:
>
> map[string]*Session
>
> https://blog.golang.org/go-maps-in-action
答案2
得分: 51
以下是Dave Cheney在《如果地图不是引用变量,那它是什么?》一文中的一些部分:
地图值是指向
runtime.hmap
结构的指针。
和结论:
结论
地图和通道一样,但与切片不同,它们只是指向运行时类型的指针。如上所示,地图只是指向
runtime.hmap
结构的指针。地图在Go程序中具有与任何其他指针值相同的指针语义。除了编译器将地图语法重写为对
runtime/hmap.go
中的函数的调用之外,没有任何魔法。
关于map
语法的历史/解释的有趣部分:
如果地图是指针,它们不应该是
*map[key]value
吗?这是一个很好的问题,如果地图是指针值,为什么表达式
make(map[int]int)
返回类型为map[int]int
的值。它不应该返回*map[int]int
吗?Ian Taylor最近在golang-nuts的线程中回答了这个问题1。在早期,我们现在称为地图的东西被写成指针,所以你写
*map[int]int
。当我们意识到没有人写map
而不写*map
时,我们就摒弃了这种写法。可以说,将类型从
*map[int]int
重命名为map[int]int
,虽然令人困惑,因为该类型看起来不像指针,但比一个无法解引用的指针形状的值更不令人困惑。
英文:
Here are some parts from If a map isn’t a reference variable, what is it? by Dave Cheney:
> A map value is a pointer to a runtime.hmap
structure.
and conclusion:
> Conclusion
>
> Maps, like channels, but unlike slices, are just pointers to runtime
> types. As you saw above, a map is just a pointer to a runtime.hmap
> structure.
>
> Maps have the same pointer semantics as any other pointer value in a
> Go program. There is no magic save the rewriting of map syntax by the
> compiler into calls to functions in runtime/hmap.go
.
And an interesting bit about history/explanation of map
syntax:
> If maps are pointers, shouldn’t they be *map[key]value
?
>
> It’s a good question that if maps are pointer values, why does the
> expression make(map[int]int)
return a value with the type
> map[int]int
. Shouldn’t it return a *map[int]int
? Ian Taylor
> answered this recently in a golang-nuts thread<sup>1</sup>.
>
> > In the very early days what we call maps now were written as pointers,
> > so you wrote *map[int]int
. We moved away from that when we realized
> > that no one ever wrote map
without writing *map
.
>
>
> Arguably renaming the type from *map[int]int
to map[int]int
, while
> confusing because the type does not look like a pointer, was less
> confusing than a pointer shaped value which cannot be dereferenced.
答案3
得分: 16
不。默认情况下,地图是引用类型。
package main
import "fmt"
func mapToAnotherFunction(m map[string]int) {
m["hello"] = 3
m["world"] = 4
m["new_word"] = 5
}
// func mapToAnotherFunctionAsRef(m *map[string]int) {
// m["hello"] = 30
// m["world"] = 40
// m["2ndFunction"] = 5
// }
func main() {
m := make(map[string]int)
m["hello"] = 1
m["world"] = 2
// 初始状态
for key, val := range m {
fmt.Println(key, "=>", val)
}
fmt.Println("-----------------------")
mapToAnotherFunction(m)
// 传递给函数作为指针后
for key, val := range m {
fmt.Println(key, "=>", val)
}
// 尝试取消注释此行
fmt.Println("-----------------------")
// mapToAnotherFunctionAsRef(&m)
// // 传递给函数作为指针后
// for key, val := range m {
// fmt.Println(key, "=>", val)
// }
// 输出
// hello => 1
// world => 2
// -----------------------
// hello => 3
// world => 4
// new_word => 5
// -----------------------
}
来自Golang博客-
> 地图类型是引用类型,就像指针或切片一样,因此上面的m的值为nil;它没有指向初始化的地图。当读取时,nil地图的行为类似于空地图,但尝试写入nil地图将导致运行时恐慌;不要这样做。要初始化地图,请使用内置的make函数:
```go
// make函数示例
m = make(map[string]int)
代码片段链接 玩一下。
英文:
No. Maps are reference by default.
package main
import "fmt"
func mapToAnotherFunction(m map[string]int) {
m["hello"] = 3
m["world"] = 4
m["new_word"] = 5
}
// func mapToAnotherFunctionAsRef(m *map[string]int) {
// m["hello"] = 30
// m["world"] = 40
// m["2ndFunction"] = 5
// }
func main() {
m := make(map[string]int)
m["hello"] = 1
m["world"] = 2
// Initial State
for key, val := range m {
fmt.Println(key, "=>", val)
}
fmt.Println("-----------------------")
mapToAnotherFunction(m)
// After Passing to the function as a pointer
for key, val := range m {
fmt.Println(key, "=>", val)
}
// Try Un Commenting This Line
fmt.Println("-----------------------")
// mapToAnotherFunctionAsRef(&m)
// // After Passing to the function as a pointer
// for key, val := range m {
// fmt.Println(key, "=>", val)
// }
// Outputs
// hello => 1
// world => 2
// -----------------------
// hello => 3
// world => 4
// new_word => 5
// -----------------------
}
From Golang Blog-
>Map types are reference types, like pointers or slices, and so the value of m above is nil; it doesn't point to an initialized map. A nil map behaves like an empty map when reading, but attempts to write to a nil map will cause a runtime panic; don't do that. To initialize a map, use the built in make function:
// Ex of make function
m = make(map[string]int)
Code Snippet Link Play with it.
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