


Why Does `json.Unmarshal` Unexpectedly Change My Struct When Using `interface{}`?
Dec 12, 2024 pm 07:53 PMMisconceptions About Go's interface{} Type
Background
The interface{} type, a significant aspect of Go programming, allows for dynamic handling of values, but its usage can often lead to misunderstandings.
An Unexpected Bug with json.Unmarshal
A common misconception arises when using interface{} as a function parameter and encountering an issue with json.Unmarshal. The following code snippet demonstrates this bug:
func test(i interface{}) { j := []byte(`{ "foo": "bar" }`) fmt.Printf("%T\n", i) fmt.Printf("%T\n", &i) json.Unmarshal(j, &i) fmt.Printf("%T\n", i) } func main() { test(Test{}) } type Test struct { Foo string }
When run, this code unexpectedly transforms the struct Test{} into a map[string]interface{} after being unmarshalled, as seen in the output:
main.Test *interface {} map[string]interface {}
Understanding the Misconception
The misunderstanding lies in the nature of the interface{} type. It's not an empty container as some may assume, but rather a wrapper for a value-type pair. When the non-pointer Test{} is passed to test(), the json.Unmarshal function expects a pointer, causing it to create a new map[string]interface{} value instead.
To resolve this issue, one must pass a pointer to the interface{} parameter.
func test(i interface{}) { j := []byte(`{ "foo": "bar" }`) fmt.Printf("%T\n", i) fmt.Printf("%T\n", &i) json.Unmarshal(j, i) fmt.Printf("%T\n", i) fmt.Println(i) } func main() { test(&Test{}) }
This modification ensures that a pointer to a pointer is passed to the function, allowing json.Unmarshal to unmarshal the JSON data correctly.
Conclusion
It's crucial to remember that interface{} requires careful usage to avoid unexpected behavior. Understanding its underlying implementation and its implications on functions like json.Unmarshal is essential for effective Go development.
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