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    <p>You have several choices. The one that makes the most sense really depends on what you're trying to do.</p> <p><strong>Choice 1: make toyNumber a public member variable in a class</strong></p> <pre><code>class MyToy { public int toyNumber; } </code></pre> <p>then pass a reference to a MyToy to your method.</p> <pre><code>void play(MyToy toy){ System.out.println("Toy number in play " + toy.toyNumber); toy.toyNumber++; System.out.println("Toy number in play after increement " + toy.toyNumber); } </code></pre> <p><strong>Choice 2: return the value instead of pass by reference</strong></p> <pre><code>int play(int toyNumber){ System.out.println("Toy number in play " + toyNumber); toyNumber++; System.out.println("Toy number in play after increement " + toyNumber); return toyNumber } </code></pre> <p>This choice would require a small change to the callsite in main so that it reads, <code>toyNumber = temp.play(toyNumber);</code>.</p> <p><strong>Choice 3: make it a class or static variable</strong></p> <p>If the two functions are methods on the same class or class instance, you could convert toyNumber into a class member variable.</p> <p><strong>Choice 4: Create a single element array of type int and pass that</strong></p> <p>This is considered a hack, but is sometimes employed to return values from inline class invocations.</p> <pre><code>void play(int [] toyNumber){ System.out.println("Toy number in play " + toyNumber[0]); toyNumber[0]++; System.out.println("Toy number in play after increement " + toyNumber[0]); } </code></pre>
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    1. POHow to do the equivalent of pass by reference for primitives in Java
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    1. COClear explanation and especially good for providing multiple choices on how to code for the desired effect. Directly helpful for something I'm working on right now! It is nuts that this question was closed.
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    2. COAlthough your choice 1 does do what you are trying to convey, I actually had to go back and double check it. The question asks if you can pass by reference; so really you should have done the printlns in the same scope that the toy passed to the function exists in. The way you have it, the printlns are operated in the same scope as the increment which doesn't really prove the point, of COURSE a local copy printed after an increment will have a different value, however that doesn't mean that the passed reference will. Again though, your example does work, but doesn't prove the point.
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    3. CONo, I think it's correct the way it is. Each function "play()" in my answer above is intended as a drop-in replacement for the original "play()" function, which also printed the value before and after incrementing. The original question has the println() in main that proves pass by reference.
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