


Stringer Interface Confusion with Println: Understanding Value-Based vs. Pointer-Based Objects
Question:
In a scenario where an object implements the Stringer interface, why doesn't the object's String method get invoked when using fmt.Println if the object is value-based?
Code Example:
Consider the following Go code:
type Car struct { year int make string } func (c *Car) String() string { return fmt.Sprintf("{make:%s, year:%d}", c.make, c.year) } func main() { myCar := Car{year: 1996, make: "Toyota"} fmt.Println(myCar) }
When myCar is a pointer, the String method gets invoked as expected. However, when myCar is a value, the default Go formatting is used instead.
Answer:
The reason for this behavior lies in the way Go interfaces work. When you specify a type that implements an interface (Stringer, in this case), Go expects that type to be the exact type of the interface. When you pass a value of type Car to fmt.Println, it is implicitly converted to interface{}, and there is no type Car in the interface{} type system. Instead, it's a type *Car (a pointer to Car).
The fmt.Println function uses a type switch to determine how to print the value based on its type. For a Stringer interface, it checks if the value implements the String method. Since Car (value-based) doesn't implement String, the default formatting is used. However, when you call myCar.String() explicitly, the compiler automatically converts it to (&myCar).String(), which has the correct *Car type and invokes the desired formatting method.
To ensure that the object is formatted as desired regardless of its type, you have two options:
- Implement String on Car (value-based): This requires creating a separate String method for Car that operates on value receivers, which would incur unnecessary object copying.
- Always pass a pointer to fmt.Println: By using fmt.Println(&myCar), you ensure that the passed value has the correct *Car type and the String method will be invoked as expected.
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