Anonymous type in method for Deserializing json from a string to DataContract

Question

I am trying to write an anonymous method for the purpose of deserializing Json to a DataContract. This would allow me to have something re-usable without having to write the same code for each DataContract class I wrote for each json query.

The code I have so far, is as follows :

public T Json2Object<T>(string json, Encoding encoding) {
    T result;
    DataContractJsonSerializer ser = new DataContractJsonSerializer( typeof( T ) );
    using ( Stream s = new MemoryStream( ( encoding ?? Encoding.UTF8 ).GetBytes( json ?? "" ) ) ) {
        result = ser.ReadObject( s ) as T;
    }
    return result;
}

It is giving me errors in the IDE as follows :

How can this be adjusted without hard-coding the type so that it works as I intend ?

Rob · Accepted Answer · 2016-12-29 04:31:14Z

1

The as keyword implies that the type is a reference type, and not a value type. You cannot store null in a value type. Thus, you either need to restrict the type T to a reference type:

public T Json2Object<T>(string json, Encoding encoding) where T : class {
    // ...
}

Or cast rather than use as:

public T Json2Object<T>(string json, Encoding encoding) {
    T result;
    DataContractJsonSerializer ser = new DataContractJsonSerializer( typeof( T ) );
    using ( Stream s = new MemoryStream( ( encoding ?? Encoding.UTF8 ).GetBytes( json ?? "" ) ) ) {
        result = (T)ser.ReadObject( s );
    }
    return result;
}

answered Dec 29, 2016 at 4:31

Rob♦

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3 Comments

Kraang Prime Over a year ago

Very nice, thank you. Where you have where T : class do I need to replace class with the specific class. For example, if my datacontract class is named Foo would this be written as where T : Foo or would that where T : class still work as you have it written ?

Rob Over a year ago

@KraangPrime Nope, where T : class is exactly what you need to write. It restricts T to a reference type. Note that you will be forbidden from writing Json2Object<int>(someJson, someEncoding) since int is a value type. If you want to be able to pass in int, you'll need to use the casting approach instead.

Kraang Prime Over a year ago

That is awesome. Exactly what I needed. Thank you :)

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