How to Override Embedded Struct Methods in Go?

Overriding Embedded Struct's Method in Go

In Go, when a struct inherits from another struct (known as embedding), it inherits all the fields and methods of the embedded struct. However, if the inheriting struct defines its own method with the same name as one in the embedded struct, the embedded method is hidden, or rather, shadowed.

Overriding the Embedded Method

Overriding a method in the embedded struct involves explicitly accessing the embedded struct's method using the syntax:

<code class="go">yourStruct.EmbeddedStruct.method()</code>

For example:

<code class="go">type Base struct {
    val int
}
func (b *Base)Set(i int) {
    b.val = i
}

type Sub struct {
    Base
    changed bool
}</code>

Here, Sub embeds Base. However, Sub's Set method overrides Base's Set method. To call Base's original Set method from within Sub, one would need to explicitly call b.Base.Set(i).

Example: Overriding and Invoking Embedded Method

Consider the following code:

<code class="go">func t16() {
    s := &amp;Sub{}
    s.Set(1)
    var b *Base = &amp;s.Base
    fmt.Printf("%+v\n", b)
    fmt.Printf("%+v\n", s)
}</code>

Initially, Sub's Set method is invoked, which overrides Base's Set method. To demonstrate the original behavior of Base's Set method, one can call s.Base.Set(10):

<code class="go">func t16() {
    s := &amp;Sub{}
    s.Base.Set(10)
    var b *Base = &amp;s.Base
    fmt.Printf("%+v\n", b)
    fmt.Printf("%+v\n", s)
}</code>

This ensures that Base's original Set method is invoked, bypassing Sub's custom Set method and its subsequent modification of changed.

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