How does Go handle methods and pointers when the receiver type and the value type don't match?

Pointers in Golang: A Comprehensive Explanation

In the Tour of Go, it is mentioned that pointers are used to implement methods on values. However, what happens if we declare a method with a non-pointer receiver but attempt to use it on a pointer value, or vice versa?

The answer lies in two fundamental rules of the Go language:

Method Derivation:
Go allows methods to be derived from existing methods. In your example, the method func (v Vertex) Abs() float64 is derived from the method func (v *Vertex) Abs() float64. This means that when you declare the former method, a new implementation is automatically generated, which essentially calls the original method with a pointer to the receiver:

func (v *Vertex) Abs() float64 { return math.Sqrt(v.X*v.X+v.Y*v.Y) }
func (v Vertex) Abs() float64 { return Vertex.Abs(*v) } // Automatically generated

Therefore, regardless of whether you declare func (v *Vertex) Abs() or func (v Vertex) Abs(), in both cases, the generated method is called.

Automatic Address Taking:
Go can automatically generate a pointer to a variable. In the following case, the expression v.Abs() resolves to the code:

vp := &v
vp.Abs()

This means that when you pass a value to a method that expects a pointer and Go cannot automatically derive the method from an existing one, the compiler will generate a pointer to the value and then call the method on the pointer.

In summary, Go follows these rules to handle pointers and methods:

• Methods with non-pointer receivers can be derived from methods with pointer receivers, and the derived method will call the original method with a pointer to the receiver.
• If Go cannot automatically derive a method from an existing one, it will generate a pointer to the value and call the method on the pointer.

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