Is is possible to remove generic type parameters from object constructor in Java?

Question

In Java it is tiring to have to write:

Pair<String, String> pair = new Pair<String, String>("one", "two");

It would be nice if the types were inferred for you so you could at least do this:

Pair<String, String> pair = new Pair("one", "two");

And skip the generic params again.

You can make a static method that can cheat around it like so:

public static <T, S> Pair<T, S> new_(T one, S two) {
    return new Pair<T, S>(one, two);
}

And then use it like: Pair.new_("one", "two").

Is it possible to build the type inferencing into the constructor so that hack can be avoided?

I was thinking of something like:

public <S,T> Pair(S one, T two) {
    this.one = one;
    this.two = two;
}

But then you run into generic type collisions. Does anyone have any thoughts?

That default behavior was actually proposed for JDK 1.7, but I don't think it got included. — JustinKSU, Jun 13 '11 at 16:26
@Yaneeve, could you please elaborate on that question? I don't know what you mean. — jjnguy, Jun 13 '11 at 16:29
@JustinKSU It was included in the form of [diamond operator](http://stackoverflow.com/questions/4166966/what-is-the-point-of-the-diamond-operator-in-java-7) — axtavt, Jun 13 '11 at 16:31
@jjnguy, I meant that when I declare a class such as Pair, and then when I wish to write a new statement, my IDE (intellij, currently, but I remember eclipse doing the same) can complete the statement (after writing the new keyword) with Pair(), after which I just add "one" then "two"... — Yaneeve, Jun 13 '11 at 16:38
@Yaneeve, good point. My IDE does auto-complete that. However, it is also more clutter to read. Esp. if you have things like `Map>`. I guess I'd like it better for readability and maintainability in the future. — jjnguy, Jun 13 '11 at 16:45
Hmm, where did they hide the language specification? http://java.sun.com/docs/books/jls/ does not work anymore, as well as the web site feedback link there. :-/ — Paŭlo Ebermann, Jun 13 '11 at 17:03
Incidentally, [beware of failure to infer wildcards](http://stackoverflow.com/q/1294227/1968) in this code. — Konrad Rudolph, Jun 13 '11 at 17:57
On a completely off-topic note, is [this](http://stackoverflow.com/questions/3389264/how-to-get-the-separate-digits-of-an-int-number/4924618#4924618) really you? — Michael Myers, Nov 09 '11 at 21:33
@Michael, I'm not sure. I never use `toCharArray` in my code...so I'm not sure why I would suggest that. I don't remember writing that, or answering a question I already had the top answer on with a completely different answer. I'm a bit spooked. — jjnguy, Nov 09 '11 at 22:54
Oh, waffles must have merged the users before you saw it. The user there was called NotJustin and had the same gravatar and a remarkably similar email. The weird thing is that according to your user history, you weren't even going by Justin on February 7. — Michael Myers, Nov 09 '11 at 23:00
No, just the one. Doesn't matter now, I guess. Just curious. — Michael Myers, Nov 10 '11 at 00:08

score 8 · Accepted Answer · answered Jun 13 '11 at 16:26

8

It s common to have a helper method which will imply the types for you.

Pair<String, String> pair = Pair.of("one", "two");

then it doesn't seem like such a hack.

It would also be nice if Java had a built in Pair class. ;)

answered Jun 13 '11 at 16:26

Peter Lawrey

525,659
79
751
1,130

1

Yeah, I noted that in my question. I was wondering if there was a way around needing that. – jjnguy Jun 13 '11 at 16:27
All you can do is hide the ugliness with a nicer name and perhaps a `import static Pair.pairOf;` and use just the method name. – Peter Lawrey Jun 13 '11 at 16:29
3

I think `Pair.of` is a lot cleaner than a static imported `pairOf` method! – ColinD Jun 13 '11 at 16:40
I think "factory method" is the term people generally use for this – newacct Jun 14 '11 at 05:08
@newacct, It is a factory method, but a helper method is one which helps hide away a piece of ugliness or complexity. This is the point I wanted to emphasis. – Peter Lawrey Jun 14 '11 at 07:36
@Oscar, haha yeah. I was surprised as well. – jjnguy Jun 14 '11 at 17:46
@jjnguy, Perhaps you should have answered your own question. ;) I suggested changing the name so it appears more natural even if technically it is the same solution. – Peter Lawrey Jun 15 '11 at 05:24
@OscarRyz, I don't always understand why I get as many or as few votes as I do. – Peter Lawrey Jun 15 '11 at 11:25

score 6 · Answer 2 · answered Aug 27 '11 at 00:05

In Java 7 you can use the "diamond" operator:

List<String> strings = new ArrayList<>();

Map<String, Int> map = new HashMap<>();

Map<String, List<String>> lolmap = new HashMap<>();

Here is the JSR which defines it: JSR334
Example using the Diamond operator and some other Java7 features: New language features in Java 7

score 1 · Answer 3 · answered Jun 13 '11 at 16:32

The problem is that the declaration of a variable of a particular type, and the instantiation of an object to be referenced by that variable, are two different operations. Strings are generally hard to mistake, but let's say you were setting up pairs of numbers (maybe X-Y coordinates):

Pair<float,float> myCoords = new Pair(3,4);

This brings up a quandary; you're declaring a Pair<float, float>, but assigning an object that would probably be inferred as Pair<int,int>, unless the compiler were given enough intelligence to take the type of the variable being set into account when inferring the instantiated object's generic types (highly unlikely). Java generics, AFAIK, are not covariant, and even if they were, int does not inherit from float. I don't know of a non-duck-typed language that would handle a situation like this correctly.

Java does not support primitives as type parameters, thus it would at least have to be `Pair` and `Pair`. — Paŭlo Ebermann, Jun 13 '11 at 18:00
My apologies; I code in C# mostly, which does allow use of the primitive "aliases" for the actual class names. — KeithS, Jun 23 '11 at 21:36

score 1 · Answer 4 · answered Jun 13 '11 at 16:34

I know it's tiring but that way you ensure type safety of your code. This code is a legal Java code (despite of the warning from the compiler):

import java.util.*;

public class Main {
  public static void main(String[] args) {
    ArrayList a = new ArrayList();
    a.add(1);
    ArrayList<String> b = a;
    System.out.println(b.get(0));
  }
}

But the compiler cannot infer the type now, yet a is assignment compatible with b. Try running it, and a runtime error would occur.

Yeah, don't use raw types here. – Paŭlo Ebermann Jun 13 '11 at 18:00 — Paŭlo Ebermann, Jun 13 '11 at 18:00

score 0 · Answer 5 · answered Jun 13 '11 at 16:29

0

I prefer things to be explicit. When things are inferred, they can be inferred incorrectly, or the method can be used incorrectly without any notice because the system infers how to handle the incorrect arguments.

answered Jun 13 '11 at 16:29

Larry Watanabe

10,126
9
43
46

Is is possible to remove generic type parameters from object constructor in Java?

5 Answers5