英文:
How cap() of slice calculated?
问题
在下面的代码中：
package main
import "fmt"
func main() {
b := make([]int, 0, 5)
fmt.Println(len(b), cap(b), b) // 0 5
c := b[:3]
fmt.Println(len(c), cap(c), c) // 3 5
d := c[1:5]
fmt.Println(len(d), cap(d), d) // 4 4
e := d[0:4]
fmt.Println(len(e), cap(e), e) // 4 4
}
d的底层数组与b和c的底层数组相同。
为什么cap(d)是4？
英文:
In the below code:
package main
import "fmt"
func main() {
b := make([]int, 0, 5)
fmt.Println(len(b), cap(b), b) // 0 5
c := b[:3]
fmt.Println(len(c), cap(c), c) // 3 5
d := c[1:5]
fmt.Println(len(d), cap(d), d) // 4 4
e := d[0:4]
fmt.Println(len(e), cap(e), e) // 4 4
}
underlying array for d is same as underlying array for b & c
Why cap(d) is 4?
答案1
得分: 1
让我们一步一步来解释。
初始化 b
b := make([]int, 0, 5) // 创建一个长度为0，容量为5的切片
fmt.Println(len(b), cap(b), b) // 输出 0 5
b => c
c := b[:3] // 创建一个长度为3，容量仍为5的切片
fmt.Println(len(c), cap(c), c) // 输出 3 5
c => d
d := c[1:5] // 创建一个长度为4，容量为4的切片
fmt.Println(len(d), cap(d), d) // 输出 4 4
c[1:5] 之所以使容量减少了一个，是因为它实际上从数组中删除了 c[0]... 它被完全切片掉了。
可视化
数组长度为5
-------------
[ 0 0 0 0 0 ]
0 1 2 3 4
c[1:5] = [ x | 0 0 0 0 ]
^
这个索引在切片后面，被切片掉了，所以新的容量是基于数组 [ 0 0 0 0 ]
1 2 3 4
为什么其他情况没有发生这种情况呢...?
-------------------------------------------
b[:3] = [ 0 0 0 | x x ]
^ ^
这两个索引没有落在切片后面，所以容量仍然是5 [ 0 0 0 x x ]
1 2 3 4 5
英文:
lets break it down step by step
intialized b
b := make([]int, 0, 5) // which makes [ ] = length 0, with a cap of 5
fmt.Println(len(b), cap(b), b) // 0 5
b => c
c := b[:3] // which makes [ 0 0 0 ] = length 3, still a cap of 5
fmt.Println(len(c), cap(c), c) // 3 5
c => d
d := c[1:5] // which makes [ 0 0 0 0 ] = length of 4, now with a cap of 4
fmt.Println(len(d), cap(d), d) // 4 4
the reason for c[1:5] making the cap one less because it's technically erasing c[0] from the array... it's being completely sliced out of it.
visualization
array of 5
-------------
[ 0 0 0 0 0 ]
0 1 2 3 4
c[1:5] = [ x | 0 0 0 0 ]
^
this index of the array fell behind the sliced indexs and was
sliced out making the new cap is based off the array [ 0 0 0 0 ]
1 2 3 4
why didnt this happen with the others...?
-------------------------------------------
b[:3] = [ 0 0 0 | x x ]
^ ^
these two indexs of the array did not fall behind the
sliced indexs which means the cap remains at 5 [ 0 0 0 x x ]
1 2 3 4 5
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