英文:
Golang: interface func to print memory address
问题
我很好奇为什么直接在变量上打印内存地址可以正常工作，但是通过接口进行相同操作时，却无法打印出内存地址。
package main
import "fmt"
type address struct {
a int
}
type this interface {
memory()
}
func (ad address) memory() {
fmt.Println("a - ", ad)
fmt.Println("a's memory address --> ", &ad)
}
func main() {
ad := 43
fmt.Println("a - ", ad)
fmt.Println("a's memory address --> ", &ad)
// 在这里初始化代码
thisAddress := address{
a: 42,
}
// 不确定为什么这里也没有返回内存地址？
var i this
i = thisAddress
i.memory()
}
我只返回翻译好的部分，不包括其他内容。
英文:
I am curious as to why just printing the memory address on a var directly works but trying to do the same action through an interface doesn't print out the memory address?
package main
import "fmt"
type address struct {
a int
}
type this interface {
memory()
}
func (ad address) memory() {
fmt.Println("a - ", ad)
fmt.Println("a's memory address --> ", &ad)
}
func main() {
ad := 43
fmt.Println("a - ", ad)
fmt.Println("a's memory address --> ", &ad)
//code init in here
thisAddress := address{
a: 42,
}
// not sure why this doesnt return memory address as well?
var i this
i = thisAddress
i.memory()
}
https://play.golang.org/p/Ko8sEVfehv
Just wanted to add this after fixing errors, it now functions as expected.
Testing shifting memory pointers
package main
import "fmt"
type address struct {
a int
}
type this interface {
memory() *int
}
func (ad address) memory() *int {
/*reflect.ValueOf(&ad).Pointer() research laws of reflection */
var b = &ad.a
return b
}
func main() {
thisAddress := address{
a: 42,
}
thatAddress := address{
a: 43,
}
var i this
i = thisAddress
a := i.memory()
fmt.Println("I am retruned", a)
fmt.Println("I am retruned", *a)
i = thatAddress
c := i.memory()
fmt.Println("I am retruned", c)
fmt.Println("I am retruned", *c)
}
答案1
得分: 2
因为在memory()方法的第二个案例中：
func (ad address) memory() {
fmt.Println("a - ", ad)
fmt.Println("a's memory address --> ", &ad)
}
ad不是一个int，而是一个结构体，ad的类型是address。你打印的不是一个int的地址，而是一个结构体的地址。指向结构体的指针的默认格式是：&{}。
引用fmt包文档中关于默认格式的说明：
> 结构体: {字段0 字段1 ...}
> 数组、切片: [元素0 元素1 ...]
> 映射: map[键1:值1 键2:值2]
> 指向上述类型的指针: &{}, &[], &map[]
如果你修改这一行代码，打印address.a字段的地址，该字段的类型是int：
fmt.Println("a's memory address --> ", &ad.a)
你将看到相同的指针格式以十六进制格式打印出来，例如：
a's memory address --> 0x1040e13c
英文:
Because in your second case inside the memory() method:
func (ad address) memory() {
fmt.Println("a - ", ad)
fmt.Println("a's memory address --> ", &ad)
}
ad is not an int but a struct, ad is of type address. And you're not printing the address of an int but an address of a struct. And the default formatting for pointer to a struct is: &{}.
Quoting from package doc of fmt regarding default formats:
> struct: {field0 field1 ...}
> array, slice: [elem0 elem1 ...]
> maps: map[key1:value1 key2:value2]
> pointer to above: &{}, &[], &map[]
If you modify the line to print the address of the address.a field which is of type int:
fmt.Println("a's memory address --> ", &ad.a)
You will see the same pointer format printed in hexadecimal format, e.g.:
a's memory address --> 0x1040e13c
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