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为什么map在Go上有不同的行为？关于在Go上使用等号和地图的说明
在Go所有类型由值复制：string，intxx，uintxx，floatxx，struct，[...]array，[]slice除了map[key]value
package main
import "fmt"
type test1 map[string]int
func (t test1) DoSomething() { // doesn't need to use pointer
t["yay"] = 1
}
type test2 []int
func (t* test2) DoSomething() { // must use pointer so changes would effect
*t = append(*t,1)
}
type test3 struct{
a string
b int
}
func (t* test3) DoSomething() { // must use pointer so changes would effect
t.a = "aaa"
t.b = 123
}
func main() {
t1 := test1{}
u1 := t1
u1.DoSomething()
fmt.Println("u1",u1)
fmt.Println("t1",t1)
t2 := test2{}
u2 := t2
u2.DoSomething()
fmt.Println("u2",u2)
fmt.Println("t2",t2)
t3 := test3{}
u3 := t3
u3.DoSomething()
fmt.Println("u3",u3)
fmt.Println("t3",t3)
}
而传递变量作为函数的参数/参数等于与:=分配
package main
import "fmt"
type test1 map[string]int
func DoSomething1(t test1) { // doesn't need to use pointer
t["yay"] = 1
}
type test2 []int
func DoSomething2(t *test2) { // must use pointer so changes would effect
*t = append(*t,1)
}
type test3 struct{
a string
b int
}
func DoSomething3(t *test3) { // must use pointer so changes would effect
t.a = "aaa"
t.b = 123
}
func main() {
t1 := test1{}
DoSomething1(t1)
fmt.Println("t1",t1)
t2 := test2{}
DoSomething2(&t2)
fmt.Println("t2",t2)
t3 := test3{}
DoSomething3(&t3)
fmt.Println("t3",t3)
}