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copied!<h2>Simple answer</h2> <p>Yes.</p> <h2>More detailed answer</h2> <p>You can either use generic collection, without <code><T></code> value, for example:</p> <pre><code>ArrayList a = new ArrayList(); a.add(2); a.add("String"); </code></pre> <p>Using collections without <code><T></code> is a bad habit and most IDEs / compilers give a warning here. You can circumvent it by using a collection of <code>Object</code>, i.e.:</p> <pre><code>ArrayList<Object> a = new ArrayList<Object>(); </code></pre> <p>Or you can find some common interface or supertype that these element must have in, for example <code>ArrayList<Number></code> - and you can store various objects that have common <code>Number</code> superclass, i.e. BigDecimal, BigInteger, Byte, Double, Float, Integer, Long, Short:</p> <pre><code>ArrayList<Number> a = new ArrayList<Number>(); a.add(2); // integer a.add(42L); // long a.add(123.45d); // double System.out.println(a.toString()); // => [2, 42, 123.45] </code></pre> <p>Note that it essentially means that <code>a</code> elements are of <code>Number</code> class — i.e. you can't ask to execute subclass-specific methods (for example, <a href="https://docs.oracle.com/javase/7/docs/api/java/lang/Double.html#isInfinite()" rel="nofollow noreferrer">Double#isInfinite()</a>, which doesn't exist in <code>Number</code> superclass), although you can typecast in run-time if you somehow know it's safe to typecast:</p> <pre><code>a.get(2).isInfinite() // compile-time error ((Double) a.get(2)).isInfinite() // => false ((Double) a.get(1)).isInfinite() // run-time error (ClassCastException) </code></pre> <p>Run-time typecasting is also generally frowned upon, as it effectively circumvents proper compile-time type safety.</p> <p>Also note that it's impossible to assign (or use) <code>ArrayList<Number></code> in place of <code>ArrayList<Integer></code> and vice-versa, i.e. this will fail to compile:</p> <pre><code>public void printNumbers(ArrayList<Number> list) { list.forEach(System.out::println); } ArrayList<Integer> a = new ArrayList<Integer>(); printNumbers(a); // "incompatible types" </code></pre> <p>as well as this:</p> <pre><code>public void printIntegers(ArrayList<Integer> list) { list.forEach(System.out::println); } ArrayList<Number> a = new ArrayList<Number>(); printIntegers(a); // "incompatible types" </code></pre> <p>To declare a variable to be able to accept both <code>ArrayList<Number></code> or any of its subclasses, one can use <code>ArrayList<? extends Number></code> or <code>ArrayList<? super Number></code> syntax. <code>extends</code> is generally used when you're going to consume (i.e. read) from the object in your method, <code>super</code> is used when you're going to produce (i.e. write). Given that printout is consuming, it's safe to use <code>extends</code>:</p> <pre><code>public void printNumbers(ArrayList<? extends Number> list) { list.forEach(System.out::println); } ArrayList<Integer> listInt = new ArrayList<Integer>(); printNumbers(listInt); // works ArrayList<Double> listDbl = new ArrayList<Double>(); printNumbers(listDbl); // also works </code></pre> <p>There is a good answer in <a href="https://stackoverflow.com/questions/4343202/difference-between-super-t-and-extends-t-in-java">Difference between <? super T> and <? extends T> in Java</a> for more in-depth explanation.</p>

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