英文:
Can you explain the behavior of this Go pointer manipulation?
问题
package main
import "fmt"
type Item struct {
val int
}
func main() {
var items []*Item
item := Item{}
items = append(items, &item)
x := items[0]
y := *x
x.val++
fmt.Printf("x=%v, y=%v\n", *x, y)
}
这段代码输出：
x={1}, y={0}
我不明白为什么这两个值不同。x 是指向数组中第一个元素的指针，我们使用 x 增加了 val 字段的值，所以第一个元素已经改变了。y 是第一个元素，它的 val 也应该改变，但是没有改变？然而，如果将 y := *x 语句移到 x.val++ 之后，那么这两个值就相等了。为什么会这样呢？
英文:
package main
import "fmt"
type Item struct {
val int
}
func main() {
var items []*Item
item := Item{}
items = append(items, &item)
x := items[0]
y := *x
x.val++
fmt.Printf("x=%v, y=%v\n", *x, y)
}
This prints:
x={1}, y={0}
I can't understand why the values are different. x is a pointer to the 1st element in the array, and we increment the val field using x, then the 1st element has been changed. y is the first element, and its val should've changed too, but didn't? If, however, the y := *x statement is moved to after x.val++, then the values are equal. Why?
答案1
得分: 3
只有这一行需要解释：
y := *x
它的意思是从指针 *x 中取值，并将其赋给 y，现在 y 有一个与 x 没有关联的新值。
英文:
Only this line needs to be explained
y := *x
and it means to take value from this pointer (*x) and assign it to y, now y has a freshly non connected to x value.
答案2
得分: -2
因为x已经是指向Item的指针。尝试使用y := x。
英文:
Because x is already a pointer to a Item. Try just y := x
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