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copied!<p>A <code>while</code> applies to the expression or block after the <code>while</code>.</p> <p>You dont have a block, so your while ends with the expression <code>dog=al.get(i);</code></p> <pre><code>while(dog.getId()!=id && i<length) dog=al.get(i); </code></pre> <p>Everything after that happens only once.</p> <p>There's no reason to new up a Dog, as you're never using the dog you new'd up; you immediately assign a Dog from the array to your dog reference.</p> <p>And if you need to get a value for a key, you should use a Map, not an Array.</p> <p>Edit: this was donwmodded why??</p> <p>Comment from OP:</p> <blockquote> <p>One further question with regards to not having to make a new instance of a Dog. If I am just taking out copies of the objects from the array list, how can I then take it out from the array list without having an object in which I put it? I just noticed as well that I didn't bracket the while-loop.</p> </blockquote> <p>A Java reference and the object it refers to are different things. They're very much like a C++ reference and object, though a Java reference can be re-pointed like a C++ pointer.</p> <p>The upshot is that <code>Dog dog;</code> or <code>Dog dog = null</code> gives you a reference that points to no object. <code>new Dog()</code> <em>creates</em> an object that can be pointed to.</p> <p>Following that with a <code>dog = al.get(i)</code> means that the reference now points to the dog reference returned by <code>al.get(i)</code>. Understand, in Java, objects are never returned, only references to objects (which are addresses of the object in memory). </p> <p>The pointer/reference/address of the Dog you newed up is now lost, as no code refers to it, as the referent was replaced with the referent you got from <code>al.get()</code>. Eventually the Java garbage collector will destroy that object; in C++ you'd have "leaked" the memory.</p> <p>The upshot is that you do need to create a variable that can refer to a Dog; you don't need to create a Dog with <code>new</code>. </p> <p>(In truth you don't need to create a reference, as what you really ought to be doing is returning what a Map returns from its get() function. If the Map isn't parametrized on Dog, like this: <code>Map<Dog></code>, then you'll need to cast the return from get, but you won't need a reference: <code>return (Dog) map.get(id);</code> or if the Map is parameterized, <code>return map.get(id)</code>. And that one line is your whole function, and it'll be faster than iterating an array for most cases.)</p>

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