为什么切片分配后内存地址会改变

Question

有两个问题我不明白。
第一种是一个slice变量赋值给另一个变量，发现新变量的地址和变量的地址不一致。我的理解是slice共享内存，按照原理，它们的地址是一样的。

第二个是slice变量容量不足时，append操作后内存地址不变。应该按原理改，因为内存不足时会重新分配内存地址。

我会很感激你的 cmets。

var a = []int{1,2,3}
fmt.Printf("%p\n",&a)
b:=a
fmt.Printf("%p\n",&b) 1)、the first question
b=append(b,0)
fmt.Printf("%p\n",&b) 2)、the second question
fmt.Println(a)
fmt.Println(b)

运行结果是：

0xc04204c3a0
0xc04204c3e0
0xc04204c3e0
[1 2 3]
[1 2 3 0]

Answer 1

切片值包含指向后备数组的指针、长度和容量。有关详细信息，请参阅Go Slices: usage and internals。

以下是对问题中代码的一些评论：

var a = []int{1, 2, 3}

// Print address of variable a
fmt.Printf("%p\n", &a)

b := a

// Print address of variable b. The variable a and b
// have different addresses.
fmt.Printf("%p\n", &b)

b = append(b, 0)

// Print address of variable b. Append did not change
// the address of the variable b.
fmt.Printf("%p\n", &b)

打印第一个切片元素的地址以获得您期望的结果。

var a = []int{1, 2, 3}

// Print address of a's first element
fmt.Printf("%p\n", &a[0])

b := a

// Print address of b's first element. This prints
// same value as previous because a and b share a backing
// array.
fmt.Printf("%p\n", &b[0])

b = append(b, 0)

// Print address of b's first element. This prints a 
// different value from previous because append allocated
// a new backing array.
fmt.Printf("%p\n", &b[0])

Answer 2

// create a new slice struct which contains length, capacity and the underlying array. 
// len(a)=3, cap(a)=3
var a = []int{1,2,3}

// `&a` means the pointer to slice struct
fmt.Printf("%p\n",&a)

// `b` is another newly created variable of slice struct, so `&b` differs from `&a`, 
// but they share the same underlying array.
// len(b)=3, cap(b)=3
b := a

// the underlying array of `b` has been extended, and been newly allocated.
// but the pointer of `b` remains.
// len(b)=4, cap(b)=6
b = append(b, 0)

希望这些cmets可以帮助到你