Does a Go Struct Implement an Interface If a Method Parameter Implements That Interface?

Struct Does Not Implement Interface If Its Method Parameter Implements the Interface

In Go, a struct implements an interface if it implements all the methods of that interface. However, if a struct method has a parameter that implements the interface, the struct will not implement the interface.

package main

type A interface {
    Close()
}

type B interface {
    Connect() (A, error)
}

type C struct {
}

func (c *C) Close() {

}

type D struct {
}

func (d *D) Connect() (*C, error) {
    c := new(C)
    return c, nil
}

func test(b B) {
}

func main() {
    d := new(D)
    test(d)
}

In the above example, the struct D does not implement the interface B because the Connect method of D has a parameter that implements the interface A. The error message you are getting is:

cannot use d (type *D) as type B in argument to test:
*D does not implement B (wrong type for Connect method)
have Connect() (*C, error)
want Connect() (A, error)

To fix this error, you need to change the type of the parameter in the Connect method of D to A.

type D struct {
}

func (d *D) Connect() (A, error) {
    c := new(C)
    return c, nil
}

Now, the struct D will implement the interface B, and you will be able to call the test() function with a D value as an argument.

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