Clarification on code snippet

[alchemist vk]

Hi folks,

 In below code, I am invoking receiver api show() via a simple uInteger type variable instead of pointer, by which it expects to be invoked . Go being  strict with type casing and I was expecting a compiler error. But to my surprise, it compiled and executed successfully.  Can you folks please explain this behavior.

package main
import "fmt"
type uInteger int

func main() {
    var x uInteger = 10
    x.show()
}

func (x *uInteger) show() {
   fmt.Printf("In show, x = %d\n", *x)
}

Thanks in Advance,

Venkatesh

[burak serdar]

On Fri, Aug 4, 2023 at 10:33 AM alchemist vk <alchem...@gmail.com> wrote:

Hi folks,

 In below code, I am invoking receiver api show() via a simple uInteger type variable instead of pointer, by which it expects to be invoked . Go being  strict with type casing and I was expecting a compiler error. But to my surprise, it compiled and executed successfully.  Can you folks please explain this behavior.

package main
import "fmt"
type uInteger int

func main() {
    var x uInteger = 10
    x.show()

Above, the variable x is addressable, so the compiler passes &x to show(). 

To call a pointer receiver method of a variable, that variable has to be addressable (it doesn't have to be a pointer), so the compiler knows how to pass the address of it. Here's how it is described in the spec:

https://go.dev/ref/spec#Method_values

 

}

func (x *uInteger) show() {
   fmt.Printf("In show, x = %d\n", *x)
}

Thanks in Advance,

Venkatesh

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[Axel Wagner]

Another relevant section is Calls (emphasis mine):

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():

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[alchemist vk]

Thank you Axel and burak for clarification.  Now I am able to understand the syntax. Thank you again.

With regards,

Venkatesh