Why doesn\'t `fmt.Println` use a struct\'s `String()` method unless the struct and method are exported?

fmt.Println and String() Methods

Why doesn't fmt.Println utilize the String() methods of members when invoked on a struct? Let's examine the underlying mechanism.

Consider the following code:

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)
}

This code produces the expected output:

{[(addr: 0x8201dde620)] (addr: 0x8201dde620)} [bar] bar

However, you may wonder why the output is not as follows:

{[bar] bar} [bar] bar

which would display the String() representation of the members instead of memory addresses.

The reason lies in the unexported nature of both the Bar type and its String() method. In Go, unexported members and methods are accessible only within the package they are defined in. Since fmt.Println is not part of the same package, it cannot access these unexported elements. Hence, it resorts to the default behavior of printing memory addresses for non-builtin types.

To address this issue, you need to make both the Bar type and its String() method exported. Additionally, the fields in the Foo struct must also be exported for proper JSON encoding and decoding.

Here's a modified version of the code that demonstrates the correct approach:

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)
}

With these changes implemented, the output will now be:

{[bar] bar} [bar] bar

This example highlights the importance of using exported types and methods for appropriate integration with formatters like fmt.Println.

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