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Why Does `json.Unmarshal` Fail with Go's `interface{}` Type?

Patricia Arquette
Patricia ArquetteOriginal
2024-12-04 14:24:12892browse

Why Does `json.Unmarshal` Fail with Go's `interface{}` Type?

Golang interface{} Type Misunderstanding

In Go, the interface{} type allows variables to hold values of any type. However, using interface{} as a function parameter type can lead to unexpected results when working with json.Unmarshal.

Scenario 1: Non-Pointer Type with Unmarshal

When passing a non-pointer type to a function parameter of type interface{}, json.Unmarshal will not be able to unmarshal into the value. Instead, it creates a new map[string]interface{} value and stores it in the interface{}. This is because json.Unmarshal requires a pointer to unmarshal into.

Scenario 2: Pointer Type with Unmarshal

If a pointer to the value is passed instead, json.Unmarshal will work as expected and unmarshal the data into the pointed value. This is because the pointer allows the json package to modify the underlying value.

Scenario 3: Incorrect Pointer Usage

Even when using a pointer type, attempting to take its address while passing to json.Unmarshal (i.e., using &&i) can lead to unexpected behavior. In this case, json.Unmarshal will dereference the pointer, find an interface{} value containing a non-pointer, and create a new map[string]interface{} value.

Explanation

interface{} is not simply a typeless container but rather a type in itself. It wraps a value and its type. Passing a pointer to interface{} allows other packages to modify the underlying value, while passing an interface{} value directly prevents modification.

Best Practice

To avoid confusion, it is recommended to use pointers to values within interfaces rather than pointers to interfaces. This approach ensures that the json package can correctly unmarshal data into the desired type.

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