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
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//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ac := 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
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s1 := make([]int, 0, 3)for i := 0; i < cap(s1); i++ { s1 = append(s1, i)}s2 := s1for 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
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//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//0xc20801e170fmt.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jn2 = 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//0xc20801e170fmt.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
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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
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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：
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[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
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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：
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[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。
