Go-append使用方法及注意事项
语法
原型
在go/src/builtin/builtin.go中对append说明如下：
// The append built-in function appends elements to the end of a slice. If
// it has sufficient capacity, the destination is resliced to accommodate the
// new elements. If it does not, a new underlying array will be allocated.
// Append returns the updated slice. It is therefore necessary to store the
// result of append, often in the variable holding the slice itself:
// slice = append(slice, elem1, elem2)
// slice = append(slice, anotherSlice...)
// As a special case, it is legal to append a string to a byte slice, like this:
// slice = append([]byte("hello "), "world"...)
func append(slice []Type, elems ...Type) []Type
要点
- 用append把一个或多个元素添加在一个slice的后面；
- append的slice有一个underlying array，此即slice和array的关系；
- 另外slice有一个length和capability的概念；
- 如果slice还有剩余的空间，可以添加这些新元素，那么append就将新的元素放在slice后面的空余空间中；
- 如果slice的空间不足以放下新增的元素，那么就需要重现创建一个数组；这时可能是alloc、也可能是realloc的方式分配这个新的数组；
- 也就是说，这个新的slice可能和之前的slice在同一个起始地址上，也可能不是一个新的地址。——通常而言，是一个新的地址。
- 分配了新的地址之后，再把原来slice中的元素逐个拷贝到新的slice中，并返回。
类比
这个slice可以和C++的vector作类比。为此，可以参考《C++源码剖析》4.2.5 vector的构造与内存管理。
示例
代码
package main
import (
"fmt"
)
type Employee struct {
name string
city string
}
func (e *Employee)String() string {
return fmt.Sprintf("Employee[address: %p, name: %s, city: %s]", e, e.name, e.city)
}
var g_employees = []Employee{}
//var g_employees = make([]Employee, 0, 4)
func add(employee *Employee) {
g_employees = append(g_employees, *employee)
}
func debug() {
count := len(g_employees)
for i := 0; i < count; i++ {
fmt.Printf("%d: %s\n", i, &(g_employees[i]))
}
fmt.Println()
}
func main() {
one := Employee{name: "name1", city:"city1"}
two := Employee{name: "name2", city:"city2"}
three := Employee{name: "name3", city:"city3"}
four := Employee{name: "name4", city:"city4"}
add(&one)
debug()
add(&two)
debug()
add(&three)
debug()
add(&four)
debug()
}
运行结果
Case 1: var g_employees = []Employee{}
0: Employee[address: 0xc42000e100, name: name1, city: city1]
0: Employee[address: 0xc4200101c0, name: name1, city: city1]
1: Employee[address: 0xc4200101e0, name: name2, city: city2]
0: Employee[address: 0xc420086000, name: name1, city: city1]
1: Employee[address: 0xc420086020, name: name2, city: city2]
2: Employee[address: 0xc420086040, name: name3, city: city3]
0: Employee[address: 0xc420086000, name: name1, city: city1]
1: Employee[address: 0xc420086020, name: name2, city: city2]
2: Employee[address: 0xc420086040, name: name3, city: city3]
3: Employee[address: 0xc420086060, name: name4, city: city4]
Case 2: var g_employees = make([]Employee, 0, 4)
0: Employee[address: 0xc420074000, name: name1, city: city1]
0: Employee[address: 0xc420074000, name: name1, city: city1]
1: Employee[address: 0xc420074020, name: name2, city: city2]
0: Employee[address: 0xc420074000, name: name1, city: city1]
1: Employee[address: 0xc420074020, name: name2, city: city2]
2: Employee[address: 0xc420074040, name: name3, city: city3]
0: Employee[address: 0xc420074000, name: name1, city: city1]
1: Employee[address: 0xc420074020, name: name2, city: city2]
2: Employee[address: 0xc420074040, name: name3, city: city3]
3: Employee[address: 0xc420074060, name: name4, city: city4]
分析
在调试打印中，特别把对象的地址打印处理。通过输出，可以有如下结论：
- 在一个capability为0的空slice上面增加一个元素的时候，加1个、2个、3个的时候会重新分配存储空间；
- 同时伴随对象拷贝。
变化的对象与不变的指针
如果slice中存放的是对象，那么在用指针处理这些对象的时候就要特别注意。比如现在全局变量采用第一种方式：
var g_employees = []Employee{}
main()代码改为：
func main() {
one := Employee{name: "name1", city:"city1"}
two := Employee{name: "name2", city:"city2"}
three := Employee{name: "name3", city:"city3"}
four := Employee{name: "name4", city:"city4"}
add(&one)
debug()
var p *Employee
p = &(g_employees[0])
fmt.Printf("p: %s\n\n", p)
add(&two)
debug()
fmt.Printf("p: %s\n\n", p)
add(&three)
debug()
fmt.Printf("p: %s\n\n", p)
add(&four)
debug()
fmt.Printf("p: %s\n\n", p)
}
在增加了一个元素之后，就定义一个指针指向该元素。运行结果：
0: Employee[address: 0xc420074000, name: name1, city: city1]
p: Employee[address: 0xc420074000, name: name1, city: city1]
0: Employee[address: 0xc420076140, name: name1, city: city1]
1: Employee[address: 0xc420076160, name: name2, city: city2]
p: Employee[address: 0xc420074000, name: name1, city: city1]
0: Employee[address: 0xc420082000, name: name1, city: city1]
1: Employee[address: 0xc420082020, name: name2, city: city2]
2: Employee[address: 0xc420082040, name: name3, city: city3]
p: Employee[address: 0xc420074000, name: name1, city: city1]
0: Employee[address: 0xc420082000, name: name1, city: city1]
1: Employee[address: 0xc420082020, name: name2, city: city2]
2: Employee[address: 0xc420082040, name: name3, city: city3]
3: Employee[address: 0xc420082060, name: name4, city: city4]
p: Employee[address: 0xc420074000, name: name1, city: city1]
可以看到指针指向的地址虽然一直不变，但已经不是我们所预期的对象了。
变通
现在让slice中存储对象地址，而不再是对象。
示例一
package main
import (
"fmt"
)
type Employee struct {
name string
city string
}
func (e *Employee)String() string {
return fmt.Sprintf("Employee[address: %p, name: %s, city: %s]", e, e.name, e.city)
}
var g_employees = []*Employee{}
func add(employee *Employee) {
g_employees = append(g_employees, employee)
}
func debug() {
count := len(g_employees)
for i := 0; i < count; i++ {
fmt.Printf("%d: %s\n", i, g_employees[i])
}
fmt.Println()
}
func main() {
one := Employee{name: "name1", city:"city1"}
two := Employee{name: "name2", city:"city2"}
three := Employee{name: "name3", city:"city3"}
four := Employee{name: "name4", city:"city4"}
add(&one)
debug()
var p *Employee
p = g_employees[0]
fmt.Printf("p: %s\n\n", p)
add(&two)
debug()
fmt.Printf("p: %s\n\n", p)
add(&three)
debug()
fmt.Printf("p: %s\n\n", p)
add(&four)
debug()
fmt.Printf("p: %s\n\n", p)
}
运行结果：
0: Employee[address: 0xc42000e1a0, name: name1, city: city1]
p: Employee[address: 0xc42000e1a0, name: name1, city: city1]
0: Employee[address: 0xc42000e1a0, name: name1, city: city1]
1: Employee[address: 0xc42000e1c0, name: name2, city: city2]
p: Employee[address: 0xc42000e1a0, name: name1, city: city1]
0: Employee[address: 0xc42000e1a0, name: name1, city: city1]
1: Employee[address: 0xc42000e1c0, name: name2, city: city2]
2: Employee[address: 0xc42000e1e0, name: name3, city: city3]
p: Employee[address: 0xc42000e1a0, name: name1, city: city1]
0: Employee[address: 0xc42000e1a0, name: name1, city: city1]
1: Employee[address: 0xc42000e1c0, name: name2, city: city2]
2: Employee[address: 0xc42000e1e0, name: name3, city: city3]
3: Employee[address: 0xc42000e200, name: name4, city: city4]
p: Employee[address: 0xc42000e1a0, name: name1, city: city1]
示例二
package main
import (
"fmt"
)
type Employee struct {
name string
city string
}
func (e *Employee)String() string {
return fmt.Sprintf("Employee[address: %p, name: %s, city: %s]", e, e.name, e.city)
}
var g_employees = []*Employee{}
func add(index int) {
g_employees = append(g_employees,
&Employee{
fmt.Sprintf("name%d", index),
fmt.Sprintf("city%d", index)})
}
func debug() {
count := len(g_employees)
for i := 0; i < count; i++ {
fmt.Printf("%d: %s\n", i, g_employees[i])
}
fmt.Println()
}
func main() {
add(1)
debug()
var p *Employee
p = g_employees[0]
fmt.Printf("p: %s\n\n", p)
add(2)
debug()
fmt.Printf("p: %s\n\n", p)
add(3)
debug()
fmt.Printf("p: %s\n\n", p)
add(4)
debug()
fmt.Printf("p: %s\n\n", p)
}
运行结果不变。