Why Doesn\'t fmt.Println Use a Struct\'s Member String() Methods in Go?

Mary-Kate Olsen
Release: 2024-11-23 07:47:27
Original
902 people have browsed it

Why Doesn't fmt.Println Use a Struct's Member String() Methods in Go?

Understanding fmt.Println's Behavior with Struct Members

In Go, when we call fmt.Println on a struct, we expect it to output the string representation of the struct's members using their respective String() methods. However, this is not always the case.

Consider the following code:

package main

import (
    "fmt"
)

type bar struct {
}

func (b bar) String() string {
    return "bar"
}

type foo struct {
    b []*bar
    bb *bar
}

func main() {
    f := foo{b: []*bar{&bar{}}, bb: &bar{}}
    fmt.Println(f, f.b, f.bb)
}
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In this code, we define a bar type with a String() method that returns the string "bar." We also define a foo type with fields b and bb, which are slices and pointers to bar types, respectively.

When we call fmt.Println on f, f.b, and f.bb, we get the following output:

{[0x176f44] 0x176f44} [bar] bar
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This is different from what we would expect, which is:

{[bar] bar} [bar] bar
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Reasons Behind fmt.Println's Behavior

There are a few reasons why fmt.Println does not use the String() method of members when called on a struct:

  • String() method is not exported: The String() method in the bar type is not exported because it starts with a lowercase letter. Exported types and methods can be accessed from other packages, while unexported ones cannot.
  • Fields are unexported: The fields b and bb in the foo type are also unexported because they start with lowercase letters. Therefore, they cannot be accessed from other packages and are not printed by fmt.Println.

Solution

To fix this, we need to make sure that the String() method and the fields in the struct are exported. Here's the corrected code:

package main

import (
    "fmt"
)

type Bar struct {
}

func (b Bar) String() string {
    return "bar"
}

type Foo struct {
    B  []Bar
    BB Bar
}

func main() {
    f := Foo{B: []Bar{Bar{}}, BB: Bar{}}
    fmt.Println(f)
}
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Now, when we run the code, we get the expected output:

{[bar] bar} [bar] bar
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