Type, *Type, Receivers & Interfaces

I’ve recently seen a bit of confusion around types, pointers to types, receivers for the two and interfaces. I’m going to try to clarify that confusion a bit with some examples.

Given a simple interface named Fooer, a struct type (Bar) that implements it and a simple function (DoFoo()) that accepts a Fooer and calls the interface function like so:

When you run the code above you get Fooer: 5 for output.

What’s happening?

Bar implements Foo() int so instances of it match the Fooer interface. If you change Bar{} to &Bar{} to create a pointer to a Bar, the program runs and still works. This is because, as per the GoSpec: pointer type *T is the set of all methods declared with receiver *T or T (that is, it also contains the method set of T). Since Foo() is declared on Bar, both an instance of Bar (Bar{}) and a pointer to an instance of Bar (&Bar{}) have the method Foo() int.

Right now Foo() isn’t doing anything interesting though, just returning the integer 5.

More interesting

Let’s modify the program a little so that Foo() actually does something: increment and return the value of a struct member.

Run the program. It may not output what you expected. You may have expected the second call to DoFoo to output Fooer: 2, but instead Fooer: 1 was output both times.

This happened because Foo() is declared on Bar (the non pointer version) and Foo()’s receiver is a copy of the Bar that was passed to DoFoo.

All parameters, including receivers, are pass by value and are copies of the original (this includes pointers btw, they are just copies of the pointer itself). Since it’s a copy, we always start out with the initial value of val, which is 0.

So would changing Bar{} to &Bar{} fix the problem? No, because even though *Bar contains all of the methods defined on Bar, they’re still called with a copy of the Bar instance since they are declared as being on Bar.

To fix, we need to declare both Foo() on *Bar AND pass a *Bar to DoFoo. With this change an instance of Bar no longer matches the Fooer interface, but a pointer to Bar (*Bar) does and works like so:


I hope these examples help clear up confusion and show the relationships between T, *T, receivers and interfaces.

Edward Muller

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