英文:
subset of array in golang
问题
我正在尝试用Python编写与以下代码行等效的代码：
H = [e for e in G if condition(e)]
以下是我的示例代码。基本上，我只是尝试使用一个getter函数（G.get）来获取G的子集。所以我认为我希望arr2是一个新的数组，但包含相同的对象。
class Object:
def __init__(self, x):
self.x = x
class Group:
def __init__(self):
self.objects = []
def get(self):
return [obj for obj in self.objects]
arr = Group()
arr.objects.append(Object(1))
print(arr.objects)
arr2 = arr.get()
arr.objects[0].x = 3
print(arr.objects)
print(arr2)
这段代码可以编译和运行，并给出以下结果：
[<__main__.Object object at 0x7f8e3e7e8a90>]
[<__main__.Object object at 0x7f8e3e7e8a90>]
[<__main__.Object object at 0x7f8e3e7e8a90>]
我的问题是，“为什么arr2不包含与arr相同的Object实例？”我相信我理解make只实例化一个Group对象，这意味着它包含一个Object。但是，for循环不应该创建一个新的Object，对吗？
谢谢帮助！
英文:
I am trying to write the equivalent of the following line in python
H = [e for e in G if condition(e)]
Here is my example code. Basically I am just trying to use a getter function (G.get) to get a subset of G. So I think I want arr2 to be a new array but containing the same object.
package main
import "fmt"
type Object struct {
x int
}
type Group []Object
func (G *Group) get() (H []Object) {
for _,v := range *G {
H = append(H,v)
}
return
}
func main() {
arr := make(Group,1)
arr[0].x = 1
fmt.Println(arr)
arr2 := arr.get()
arr[0].x = 3
fmt.Println(arr)
fmt.Println(arr2)
}
Which compiles and runs and gives me
[{1}]
[{3}]
[{1}]
My question is "Why does arr2 not contain the same instance of an Object as arr?" I believe I understand make only instantiates a Group thing which means it includes one Object in it. But then the for loop shouldn't create a new Object should it?
Thanks for the help!
答案1
得分: 2
这段简单的代码片段展示了发生了什么：
var a Object
a.x = 1
b := a
fmt.Println(a, b) // 输出 {1} {1}
b.x = 2
fmt.Println(a, b) // 输出 {1} {2}
语句 b := a 将变量 a 中的 Object 值复制到变量 b 中。现在有两个值的副本。更改其中一个不会影响另一个。
这与 Python 不同，Python 中的赋值是复制一个对值的引用。
问题中的循环也复制了值：
for _,v := range *G {
H = append(H,v)
}
变量 v 是从 *G 的切片元素的副本。v 的副本被追加到切片 H 中。有三个独立的 Object 值：*G 切片中的值，v 中的值和 H 切片中的值。
如果你希望所有的切片都引用同一个 Object 值，那么在切片中存储指向 Object 值的指针：
type Group []*Object
arr := make(Group, 1)
arr[0] = &Object{x: 1}
fmt.Println(arr[0].x) // 输出 1
arr2 := arr.get()
arr[0].x = 3
fmt.Println(arr[0].x) // 输出 3
fmt.Println(arr2[0].x) // 输出 3
英文:
This simpler snippet of code shows what's going on:
var a Object
a.x = 1
b := a
fmt.Println(a, b) // prints {1} {1}
b.x = 2
fmt.Println(a, b) // prints {1} {2}
The statement b := a copies the Object value in variable a to the variable b. There are now two copies of the value. Changing one does not change the other.
This is different from Python where assignment copies a reference to a value.
The loop in the question also copies the values:
for _,v := range *G {
H = append(H,v)
}
The variable v is a copy the slice element from *G. A copy of v is appended to slice H. There are three independent Object values: the value in slice *G, the value in v and the value in slice H.
If you want the slices to all refer to the same Object value, then store pointers to Object values in the slice:
type Group []*Object
arr := make(Group, 1)
arr[0] = &Object{x: 1}
fmt.Println(arr[0].x) // prints 1
arr2 := arr.get()
arr[0].x = 3
fmt.Println(arr[0].x) // prints 3
fmt.Println(arr2[0].x) // prints 3
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