Why Does a Go Method with a Pointer Receiver Modify a Non-Pointer Argument?

Why Method with Pointer Receiver Still Changes Value when Receives Non-Pointer?

In Go, methods with pointer receivers are known to preserve the original values of variables they operate on. However, there seems to be an apparent contradiction when a method expects a pointer argument and is passed a non-pointer value.

Consider the below snippet for Exercise 51 in the Tour of Go:

func (v *Vertex) Scale(f float64) {
    v.X *= f
    v.Y *= f
}

func main() {
    v := &Vertex{3, 4}
    v.Scale(2)
    fmt.Println(v) // Output: &{6 8}
}

According to the explanation, the Scale method should not modify v since it receives a pointer to a Vertex (*Vertex) instead of a Vertex itself.

However, when we change the v := &Vertex{3, 4} line to v := Vertex{3, 4} in main, the output changes from &{6 8} to {6 8} (notice the missing &). This indicates that v has been modified, even though it was passed as a non-pointer value.

The Explanation

The key lies in Go's strongly typed nature and compiler optimizations. Even though the Scale method expects a pointer receiver, it can still accept non-pointer values because Go implicitly converts them to pointers.

Specifically, the compiler rewrites the following code:

v := Vertex{3, 4}
v.Scale(2)

to:

(&v).Scale(2)

This means the method is actually called with a pointer to v, which allows it to modify the original value.

The following Go-specific rule explains this behavior:

A method call x.m() is valid if the method set of (the type of) x contains m and the argument list can be assigned to the parameter list of m. If x is addressable and &x's method set contains m, x.m() is shorthand for (&x).m().

In summary, methods with pointer receivers can still modify non-pointer variables because the compiler automatically converts them to pointers to preserve method type safety.

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