Diamond and Generic Methods

[Dan Smith]

Paul,

I'm not quite sure what your concern is.  But, whatever it is, it is orthogonal to diamond.  As Maurizio's example demonstrates, any instance creation expression using diamond ('new Foo<>') can be re-expressed as an invocation of a generic method ('makeFoo()').  Inference in both cases behaves the same.

Taking a stab at it, I think you're saying inference chooses 'T' as the type argument for 'new ArrayList<>' (or has "the appearance" of choosing 'T').  This is incorrect.  Inference chooses 'Object' as the type argument.

—Dan

On May 5, 2011, at 8:34 AM, Paul Benedict wrote:

> That's interesting. Because inference is supposed to be a help to
> explicit parametrization....
> 
> This would be obviously rejected by the compiler:
> identity(new java.util.ArrayList<T>());
> 
> Yet, as the compiler knows, the inference says T is Object so use Object.
> 
> While totally understandable, I also think that will be a learning
> curve for some people. Java 5 taught lots of people that you can't
> instantiate using <T> but the new Java 7 code can give the appearance
> you suddenly can.
> 
> Paul
> 
> On Thu, May 5, 2011 at 10:18 AM, Maurizio Cimadamore
> <maurizio.cimadamore at oracle.com> wrote:
>> On 05/05/11 16:06, Jesper Öqvist wrote:
>>> So, after searching the archives and scouring the JSR334 v0.875 draft, I
>>> have a question about Diamond. While the specification draft does not
>>> explicitly disallow it, it seems weird that this works (Java7 developer
>>> preview b139):
>>> 
>>> class C {
>>> <T>  void identity(T a) { return a; }
>>>     void m() {
>>>       identity(new java.util.ArrayList<>());
>>>     }
>>> }
>>> 
>>> Since there is a distinction between raw types and parameterized types
>>> with inferred type arguments (in the discussion part of the Diamond
>>> section in JSR334), it would be interesting to know what type is
>>> instantiated in the above example.
>> Hi
>> The above code is equivalent to the following:
>> 
>> class C {
>> static<Z>  ArrayList<Z>  makeArrayList() { return new ArrayList<Z>(); }
>> 
>> <T>  void identity(T a) { return a; }
>>    void m() {
>>      identity(makeArrayList());
>>    }
>> }
>> 
>> In both cases, an ArrayList<Object> is created - note that, because of
>> the restrictions in Java method type-inference, the information on
>> formal argument type is not used to infer a better type. In this
>> particular case it might not seem a big problem, but in the following
>> case you  get a compile-time error:
>> 
>> class C {
>> static<Z>  ArrayList<Z>  makeArrayList() { return new ArrayList<Z>(); }
>> 
>> <T>  T identity(T a) { return a; }
>>    void m() {
>>      //error - ArrayList<Object>  found - expected ArrayList<String>
>>      ArrayList<String>  s = identity(makeArrayList());
>>    }
>> }
>> 
>> 
>> Maurizio
>> 
>>> /Jesper
>>> 
>> 
>> 
>> 
>