英文:
Why appending on slice modified another slice?
问题
package main
import "fmt"
func main() {
src := []int{0, 1, 2, 3, 4, 5, 6}
a := src[:3]
b := src[3:]
a = append(a, 9)
fmt.Println(a, b)
}
输出结果：
[0 1 2 9] [9 4 5 6]
append函数是否修改了底层数组，使其变为 []int{0, 1, 2, 9, 4, 5, 6}？
切片a被复制为一个新的切片，具有新的底层数组，值为[0, 1, 2, 9]，而切片b仍然指向被修改的旧数组。
谢谢任何提示，非常感谢。
英文:
package main
import "fmt"
func main() {
src := []int{0, 1, 2, 3, 4, 5, 6}
a := src[:3]
b := src[3:]
a = append(a, 9)
fmt.Println(a, b)
}
output:
> [0 1 2 9] [9 4 5 6]
Did append modified the underlay array as []int{0, 1, 2, 9, 4, 5, 6}?
Slice a was copied as a new slice with a new underlay array with value [0, 1, 2, 9] and slice b still pointing to the old array that was modified.
Thanks for any hints, much appreciated
答案1
得分: 0
将切片a复制为一个新的切片，新的底层数组的值为[0, 1, 2, 9]，而切片b仍指向被修改的旧数组。
a := src[:3]创建了一个切片（指向src头部的指针，长度为3，容量为7）b := src[3:]创建了一个切片（指向src[3]的指针，长度为4，容量为4）a和b共享由src创建的相同内存a = append(a, 9)，当向同一切片追加元素时，只要不超过容量，就会修改同一数组
append是否修改了底层数组为 []int{0, 1, 2, 9, 4, 5, 6}？
是的
如果append超过了a的cap，将会分配一个新的数组，并将数据复制到新的数组中。
请尝试以下代码：
package main
import "fmt"
func main() {
src := []int{0, 1, 2, 3, 4, 5, 6}
a := src[:3]
b := src[3:]
a = append(a, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9)
fmt.Println(a, b)
}
输出结果：
[0 1 2 9 9 9 9 9 9 9 9 9 9] [3 4 5 6]
英文:
-
Slice a was copied as a new slice with a new underlay array with value [0, 1, 2, 9] and slice b still pointing to the old array that was modified.
> -a := src[:3]created a slice (a pointer to the src head, length=3, capacity=7)
> -b := src[3:]created a slice(a pointer to the src[3],length=4, capacity=4)
> -aandbshares the same memory created bysrc
> -a = append(a, 9),when appending to the same slice, as long as it does not exceed cap, it is the same array that is modified -
Did append modified the underlay array as []int{0, 1, 2, 9, 4, 5, 6}
> YES
If the append exceed the cap of a, new array will be allocated and data will be copied the the new array
try this out:
package main
import "fmt"
func main() {
src := []int{0, 1, 2, 3, 4, 5, 6}
a := src[:3]
b := src[3:]
a = append(a, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9)
fmt.Println(a, b)
}
Output:
> [0 1 2 9 9 9 9 9 9 9 9 9 9] [3 4 5 6]
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