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一、类型
数组是值类型,将一个数组赋值给另一个数组时,传递的是一份拷贝。
切片是引用类型,切片包装的数组称为该切片的底层数组。
我们来看一段代码
//a是一个数组,注意数组是一个固定长度的,初始化时候必须要指定长度,不指定长度的话就是切片了 a := [3]int{1, 2, 3} //b是数组,是a的一份拷贝 b := a //c是切片,是引用类型,底层数组是a c := a[:] for i := 0; i < len(a); i++ { a[i] = a[i] + 1 } //改变a的值后,b是a的拷贝,b不变,c是引用,c的值改变 fmt.Println(a) //[2,3,4] fmt.Println(b) //[1 2 3] fmt.Println(c) //[2,3,4]
二、make
make 只能用于slice
, map
和 channel
, 所以下面一段代码生成了一个slice
,是引用类型
s1 := make([]int, 0, 3) for i := 0; i < cap(s1); i++ { s1 = append(s1, i) } s2 := s1 for i := 0; i < len(a); i++ { s1[i] = s1[i] + 1 } fmt.Println(s1) //[1 2 3] fmt.Println(s2) //[1 2 3]
三、当对slice append 超出底层数组的界限时
//n1是n2的底层数组 n1 := [3]int{1, 2, 3} n2 := n1[0:3] fmt.Println("address of items in n1: ") for i := 0; i < len(n1); i++ { fmt.Printf("%p\n", &n1[i]) } //address of items in n1: //0xc20801e160 //0xc20801e168 //0xc20801e170 fmt.Println("address of items in n2: ") for i := 0; i < len(n2); i++ { fmt.Printf("%p\n", &n2[i]) } //address of items in n2: //0xc20801e160 //0xc20801e168 //0xc20801e170 //对n2执行append操作后,n2超出了底层数组n1的j n2 = append(n2, 1) fmt.Println("address of items in n1: ") for i := 0; i < len(n1); i++ { fmt.Printf("%p\n", &n1[i]) } //address of items in n1: //0xc20801e160 //0xc20801e168 //0xc20801e170 fmt.Println("address of items in n2: ") for i := 0; i < len(n2); i++ { fmt.Printf("%p\n", &n2[i]) } //address of items in n2: //0xc20803a2d0 //0xc20803a2d8 //0xc20803a2e0 //0xc20803a2e8
四、引用“失效”
实现了删除slice
最后一个item
的函数
func rmLast(a []int) { fmt.Printf("[rmlast] the address of a is %p", a) a = a[:len(a)-1] fmt.Printf("[rmlast] after remove, the address of a is %p", a) }
调用此函数后,发现原来的slice
并没有改变
func main() { xyz := []int{1, 2, 3, 4, 5, 6, 7, 8, 9} fmt.Printf("[main] the address of xyz is %p\n", xyz) rmLast(xyz) fmt.Printf("[main] after remove, the address of xyz is %p\n", xyz) fmt.Printf("%v", xyz) //[1 2 3 4 5 6 7 8 9] }
打印出来的结果如下:
[main] the address of xyz is 0xc2080365f0 [rmlast] the address of a is 0xc2080365f0 [rmlast] after remove, the address of a is 0xc2080365f0 [main] after remove, the address of xyz is 0xc2080365f0 [1 2 3 4 5 6 7 8 9]
这里直接打印了slice
的指针值,因为slice
是引用类型,所以指针值都是相同的,我们换成打印slice
的地址看下
func rmLast(a []int) { fmt.Printf("[rmlast] the address of a is %p", &a) a = a[:len(a)-1] fmt.Printf("[rmlast] after remove, the address of a is %p", &a) } func main() { xyz := []int{1, 2, 3, 4, 5, 6, 7, 8, 9} fmt.Printf("[main] the address of xyz is %p\n", &xyz) rmLast(xyz) fmt.Printf("[main] after remove, the address of xyz is %p\n", &xyz) fmt.Printf("%v", xyz) //[1 2 3 4 5 6 7 8 9] }
结果:
[main] the address of xyz is 0xc20801e1e0 [rmlast] the address of a is 0xc20801e200 [rmlast] after remove, the address of a is 0xc20801e200 [main] after remove, the address of xyz is 0xc20801e1e0 [1 2 3 4 5 6 7 8 9]
这次可以看到slice
作为函数参数传入函数时,实际上也是拷贝了一份slice
,因为slice
本身是个指针,所以从现象来看,slice
是引用类型
总结
以上就是这篇文章的全部内容,希望对大家的学习或者工作带来一定的帮助,如果有疑问大家可以留言交流。
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