在类型结构上包装许多方法之一？

Question

A library I use has a type with multiple methods:

type Foo struct {
}

func (f *Foo) method1() int { ... }
func (f *Foo) method2() int { ... }
func (f *Foo) method3() int { ... }
// ... and so on

I'd really like to apply some specific behavior anytime I call method1 on this type:

func (f *Foo) method1Wrapper() int {
    incrementCounter()
    return f.method1()
}

But this adds a new method I'd have to call instead of just directly calling method1() itself.

Alternatively, I imagine I could create my own type:

type Foo2 struct {
    Foo
}

func (f *Foo2) method1() int {
    incrementCounter()
    return f.Foo.method1()
}

But then I'd have to create a bunch of boilerplate code to proxy all the calls to method2, method3, etc down to the Foo implementation, and I'd have to change all my usages of Foo to Foo2.

Answer 1

You don't need to proxy the method calls to delegate them, this is exactly what embedding is for.

type Foo struct {
}

func (f *Foo) method1() int { return 1 }
func (f *Foo) method2() int { return 2 }
func (f *Foo) method3() int { return 3 }

type Foo2 struct {
    *Foo
}

func (f Foo2) method1() int {
    fmt.Println("WRAPPED")
    return f.Foo.method1()
}

Foo2 has the same method set as Foo, but method1 is "intercepted" by Foo2, which explicitly calls Foo.method1 internally.

https://play.golang.org/p/q8tea5ZuHl4