英文:
Memory address for struct not visible
问题
在Go语言中，可以获取变量（如int）的内存地址，但不能获取结构体的内存地址。以一个示例来说明：
package main
import "fmt"
type stud struct {
name string
school string
}
func main() {
stud1 := stud{"name1", "school1"}
a := 10
fmt.Println("&a is:", &a)
fmt.Println("&stud1 is:", &stud1)
}
输出结果为：
&a is: 0x20818a220
&stud1 is: &{name1 school1}
为什么&a会给出内存地址，而&stud1没有给出确切的内存位置呢？我并没有打算使用内存地址，只是对这种不同的行为感到好奇。
英文:
In Go, I was confused about why the memory address for variables like int can be obtained but not for structs. As an example:
package main
import "fmt"
func main() {
stud1 := stud{"name1", "school1"}
a:=10
fmt.Println("&a is:", &a)
fmt.Println("&stud1 is:",&stud1)
}
output is:
&a is: 0x20818a220
&stud1 is: &{name1 school1}
Why is &a giving the memory address, however &stud1 not giving the exact memory location. I don't have any intention of using the memory address but just was curious about the different behavior.
答案1
得分: 7
fmt 包使用反射来打印值，并且有一个特殊情况可以将指向结构体的指针打印为 &{Field Value}。
如果你想要看到内存地址，可以使用指针格式化动词 %p。
fmt.Printf("&stud is: %p\n", &stud)
英文:
the fmt package uses reflection to print out values, and there is a specific case to print a pointer to a struct as &{Field Value}.
If you want to see the memory address, use the pointer formatting verb %p.
fmt.Printf("&stud is: %p\n", &stud)
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