How to print the slice of pointers to get the valu

问题:

This kind output required for debugging purpose. To get actual value of slice of pointers, every time a iteration is getting required.

Is there any way, we can directly have the value instead of the address of each item present at slice using simple fmt.printf()?

Here is a code snippet : https://play.golang.org/p/bQ5vWTlKZmV

package main

import (
    "fmt"
)

type user struct {
    userID int
    name   string
    email  string
}

func main() {
    var users []*user
    addUsers(users)
}

func addUsers(users []*user) {
    users = append(users, &user{userID: 1, name: "cooluser1", email: "[e-mail]"})
    users = append(users, &user{userID: 2, name: "cooluser2", email: "[e-mail]"})
    printUsers(users)
    printEachUser(users)

}

func printUsers(users []*user) {
    fmt.Printf("users at slice %v \n", users)
}

func printEachUser(users []*user) {
    for index, u := range users {
        fmt.Printf("user at user[%d] is : %v \n", index, *u)
    }
}

At above code, if I am printing the slice directly by fmt.printf , I get only the address of the values instead of actual value itself.

output : users at slice [0x442260 0x442280]

To read to the values always, i have to call func like printEachUser to iterate the slice and get the appropriate value .

output: user at user[0] is : {1 cooluser1 [e-mail]} user at user[1] is : {2 cooluser2 [e-mail]}

Is there any way, we can read the values inside the slice of pointers using fmt.printf and get value directly like below ?

users at slice [&{1 cooluser1 [e-mail]} , &{2 cooluser2 [e-mail]}]

回答1:

This kind output required for debugging purpose.

Is there any way, we can read the values inside the slice of pointers using fmt.Printf and get value directly like below ?

users []*user
fmt.Printf("users at slice %v \n", users)

users at slice [&{1 cooluser1 [e-mail]}, &{2 cooluser2 [e-mail]}]

---

Package fmt

import "fmt"

type Stringer

Stringer is implemented by any value that has a String method, which defines the “native” format for that value. The String method is used to print values passed as an operand to any format that accepts a string or to an unformatted printer such as Print.

type Stringer interface {
        String() string
}

---

For example,

package main

import (
    "fmt"
)

type user struct {
    userID int
    name   string
    email  string
}

type users []*user

func (users users) String() string {
    s := "["
    for i, user := range users {
        if i > 0 {
            s += ", "
        }
        s += fmt.Sprintf("%v", user)
    }
    return s + "]"
}

func addUsers(users users) {
    users = append(users, &user{userID: 1, name: "cooluser1", email: "[e-mail]"})
    users = append(users, &user{userID: 2, name: "cooluser2", email: "[e-mail]"})

    fmt.Printf("users at slice %v \n", users)
}

func main() {
    var users users
    addUsers(users)
}

Playground: https://play.golang.org/p/vDmdiKQOpqD

Output:

users at slice [&{1 cooluser1 [e-mail]}, &{2 cooluser2 [e-mail]}] 

回答2:

Code : https://play.golang.org/p/rBzVZlovmEc

Output :

users at slice [{1 cooluser1 [e-mail]} {2 cooluser2 [e-mail]}]

Using stringers you can achive it.

Refer: https://golang.org/pkg/fmt/#Stringer

package main

import (
    "fmt"
)

type user struct {
    userID int
    name   string
    email  string
}

func (t user) String() string {
    return fmt.Sprintf("{%v %v %v}", t.userID, t.name, t.email)
}

func main() {
    var users []*user
    addUsers(users)
}

func addUsers(users []*user) {
    users = append(users, &user{userID: 1, name: "cooluser1", email: "[e-mail]"})
    users = append(users, &user{userID: 2, name: "cooluser2", email: "[e-mail]"})
    printUsers(users)
}

func printUsers(users []*user) {
    fmt.Printf("users at slice %v \n", users)
}

You need not apply stringer to users i.e []*users instead if you do it just for a single user it'll work fine. Also it reduces down the string operations you need to do manually making your code elegant.

回答3:

You can use spew

go get -u github.com/davecgh/go-spew/spew

func spewDump(users []*user) {
    _, err := spew.Printf("%v", users)
    if err != nil {
        fmt.Println("error while spew print", err)
    }
}

Output:

[<*>{1 cooluser1 [e-mail]} <*>{2 cooluser2 [e-mail]}]