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切片重组（reslice）
我们已经知道切片创建的时候通常比相关数组小，例如：
slice1 := make([]type, start_length, capacity)
其中
start_length 作为切片初始长度而 capacity 作为相关数组的长度。这么做的好处是我们的切片在达到容量上限后可以扩容。改变切片长度的过程称之为切片重组 reslicing，做法如下：
slice1 = slice1[0:end]，其中 end 是新的末尾索引（即长度）。将切片扩展 1 位可以这么做：
sl = sl[0:len(sl)+1]
切片可以反复扩展直到占据整个相关数组。
package main
import "fmt"
func main() {
slice1 := make([]int, 0, 10)
// load the slice, cap(slice1) is 10:
for i := 0; i < cap(slice1); i++ {
slice1 = slice1[0:i+1]
slice1[i] = i
fmt.Printf("The length of slice is %d\n", len(slice1))
}
// print the slice:
for i := 0; i < len(slice1); i++ {
fmt.Printf("Slice at %d is %d\n", i, slice1[i])
}
}
输出结果：
The length of slice is 1
The length of slice is 2
The length of slice is 3
The length of slice is 4
The length of slice is 5
The length of slice is 6
The length of slice is 7
The length of slice is 8
The length of slice is 9
The length of slice is 10
Slice at 0 is 0
Slice at 1 is 1
Slice at 2 is 2
Slice at 3 is 3
Slice at 4 is 4
Slice at 5 is 5
Slice at 6 is 6
Slice at 7 is 7
Slice at 8 is 8
Slice at 9 is 9
另一个例子：
var ar = [10]int{0,1,2,3,4,5,6,7,8,9}
var a = ar[5:7] // reference to subarray {5,6} - len(a) is 2 and cap(a) is 5
将 a 重新分片：
a = a[0:4] // ref of subarray {5,6,7,8} - len(a) is now 4 but cap(a) is still 5
问题 7.7
1) 如果 a 是一个切片，那么
s[n:n] 的长度是多少？2)
s[n:n+1] 的长度又是多少？Last modified 5yr ago