Archive for the ‘scjp’ Category

h1

Not All Assignments are Created Equal

January 30, 2007

Many introductory Java books will tell you that the compound assignment operators (e.g. b += 1) are just a shorthand for doing an assignment where something is assigned the result of itself involved in some arithmetic operation (e.g. b = b + 1). In this article we’ll see that the difference between compound operators and their supposed expanded versions can be the difference between running and failing to compile.

These are supposed to be equivalent right?


byte b1 = 0;
b1 = b1 + 1;

byte b2 = 0;
b2 += 1;

Well, that’s what a lot of books will tell you. For many cases they’re very close to equivalent and it’s certainly a fair way to describe them to a junior programmer that is just being exposed to compound operators for the first time. Once you start studying for the Sun Certified Java Programmer (SCJP) test or just dig a little deeper into the language you find that there is a subtle, yet important difference.

This will not compile:


byte b1 = 0;
b1 = b1 + 1;

Why not? It looks pretty straightforward. The reason lies in how Java handles integer arithmetic. All integral expressions in Java result in at least an int. That means that if the operations involve a long they’ll result in a long, but if they involve ints, shorts or bytes (the integral types that are smaller than a long) they will still result in an int. Since an int does not fit in a byte, it fails at compile time.

On the other hand, this compiles fine:


byte b2 = 0;
b2 += 1;

So why does that work? Isn’t it pretty much the same thing as the code above that won’t compile? It turns out that Java inserts an implicit cast when the compound operator is used. The compound assignment is actually translated to something like this:


byte b2 = 0;
b2 = (byte)(b2 + 1);

The implicit cast handles the problem of narrowing that comes from assigning an int (32 bits) to a byte (8 bits).

Joshua Smith

References


SCJP Sun Certified Java Programmer Study Guide

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h1

main() Methods in Java

January 29, 2007

Introduction

The main() method is probably one of the most familar structures in the Java language. It can easily be auto-generated in Eclipse by typing “main” followed by Ctrl-Space and in NetBeans by typing “psvm” followed by a space. It’s the entry point into thousands of applications and while it’s familiar, it also has a few hidden secrets. In this article we’ll look at more than a dozen variations of the main() method. Some will compile and run as expected. Others will not compile at all. Still others will compile and run, but can’t be used as an entry point into an application.

The Methods

Look at the methods below. Which ones will not compile? Which ones will compile, but can’t be used as entry points into an application? Which ones compile and act as you would expect a main method to act?


public static void main(String[] args) {
    System.out.println("Main1!");
}

public static void main(String[] someOtherName) {
    System.out.println("Main2!");
}

public static void main(String... args) {
    System.out.println("Main3!");
}

public static void main(String[] args)
  throws Exception {
    System.out.println("Main4!");
}

static void main(String[] args) {
    System.out.println("Main5!");
}

public void main(String[] args) {
    System.out.println("Main6!");
}

public static void main(String args[]) {
    System.out.println("Main7!");
}

public static void main(String[] args[]) {
    System.out.println("Main8!");
}

public static void main(String[][] args) {
    System.out.println("Main9!");
}

public static void main(String args) {
    System.out.println("Main10!");
}

public static void main(String[] args)
  throws IOException {
    System.out.println("Main11!");
}

static public void main(String[] args) {
    System.out.println("Main12!");
}

public strictfp static void main(String[] args) {
    System.out.println("Main13!");
}

void public static main(String[] args) {
    System.out.println("Main14!");
}

public static void main(int[] args) {
    System.out.println("Main15!");
}

public static void main(String[] args) {
    System.out.println("Main16!");
}

public static void Main(String[] args) {
    System.out.println("Main17!");
}
    

The Answers


    
/**
 * Fine.
 * 
 * This is the most common form of the main method.
 */
public static void main(String[] args) {
    System.out.println("Main1!");
}

/**
 * Fine.
 * 
 * This is the most common form of the main method except
 * that the variable accepting command line arguments has
 * been renamed to someOtherName. The name of the variable
 * is insignificant.
 */
public static void main(String[] someOtherName) {
    System.out.println("Main2!");
}

/**
 * Fine.
 * 
 * Varargs form of the main method. New with Java 5.
 */
public static void main(String... args) {
    System.out.println("Main3!");
}

/**
 * Fine.
 * 
 * This is the most common form of the main method, except
 * that it throws an exception. This is completely valid.
 */
public static void main(String[] args)
  throws Exception {
    System.out.println("Main4!");
}

/**
 * Compiles, but cannot be executed from the command line.
 * 
 * Method must be public.
 * 
 * Since the signature doesn't match it's a completely
 * different method that just happens to be called main.
 * 
 */
static void main(String[] args) {
    System.out.println("Main5!");
}

/**
 * Compiles, but cannot be executed from the command line.
 * 
 * Method must be static.
 * 
 * Since the signature doesn't match it's a completely
 * different method that just happens to be called main.
 * 
 */
public void main(String[] args) {
    System.out.println("Main6!");
}

/**
 * Fine.
 * 
 * This is the most common form of the main method
 * except that the square
 * brackets for the String array have been put beside
 * the variable. This is valid, but many think, harder
 * to read.
 */
public static void main(String args[]) {
    System.out.println("Main7!");
}

/**
 * Although the syntax is strange, this compiles, but
 * cannot be executed from the command line.
 *
 * This is the most common form of the main method, but
 * the square brackets for the args array are beside the
 * type as well as beside the args variable. They should
 * be beside one or the other, not both.
 *
 * While I would have guessed that this would not
 * compile at all, it turns out that this is equivalent
 * to main taking a two-dimensional array as a parameter.
 * String[] args[] is the same as String[][] args or String
 * args[][]. While it's certainly valid for a method to
 * accept a two-dimensional array of Strings, this does
 * not fit the required signature for a main method
 * that is to be invoked from the command line. Attempting
 * to execute this will result in the following error
 * message: Exception in thread "main"
 * java.lang.NoSuchMethodError: main
 *
 */
public static void main(String[] args[]) {
    System.out.println("Main8!");
}

/**
 * Compiles, but cannot be executed from the command line.
 *
 * The main() method needs to accept an array of Strings
 * as a parameter. The method below accepts a
 * two-dimensional array of Strings.
 *
 * Since the signature doesn't match it's a completely
 * different method that just happens to be called main.
 *
 */
public static void main(String[][] args) {
    System.out.println("Main9!");
}

/**
 * Compiles, but cannot be executed from the command
 * line.
 *
 * The main() method needs to accept an array of Strings
 * as a parameter. The method below accepts a single
 * String called args.
 *
 * Since the signature doesn't match it's a completely
 * different method that just happens to be called main.
 *
 */
public static void main(String args) {
    System.out.println("Main10!");
}

/**
 * Fine.
 * 
 * Throwing a checked exception, like IOException,
 * is legal.
 * 
 */
public static void main(String[] args)
  throws IOException {
    System.out.println("Main11!");
}

/**
 * Fine.
 * 
 * This is the most common form of the main method except
 * that static and public keywords are reversed. Their
 * order does not matter.
 * 
 */
 static public void main(String[] args) {
    System.out.println("Main12!");
}

/**
 * Fine.
 * 
 * It's perfectly acceptable to have a strictfp main
 * method.
 * 
 */
public strictfp static void main(String[] args) {
    System.out.println("Main13!");
}

/**
 * Does not compile.
 * 
 * The return type (void in this case) must come
 * immediately before the method name.
 * 
 */
 void public static main(String[] args) {
    System.out.println("Main14!");
}

/**
 * Compiles, but cannot be run from the command line.
 *
 * The main() method must accept an array of Strings,
 * not ints, as a parameter.
 *
 * Since the signature doesn't match it's a completely
 * different method that just happens to be called main.
 *
 */
public static void main(int[] args) {
    System.out.println("Main15!");
}

/**
 * Fine.
 * 
 * This is the most common form of the main method
 * except that there aren't any spaces between the
 * type, the square brackets and the variable name.
 * It's still legal.
 *
 */
public static void main(String[] args) {
    System.out.println("Main16!");
}

/**
 * Compiles, but cannot be run from the command line.
 *
 * The main() method must be all lower case.
 *
 * Since the signature doesn't match it's a completely
 * different method that just happens to be called main.
 *
 */
public static void Main(String[] args) {
    System.out.println("Main17!");
}

So why would you want to know so much about the main() method anyway? Well, besides being essential if you’re taking the Sun Certified Java Programmer (SCJP) exam, there are a few alternate formulations of the method that might come in handy. Plus, with the examples above it mind, I bet you’ll spot a flawed main() method quicker than most if you come across one.

Joshua Smith


References


SCJP Sun Certified Java Programmer Study Guide

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h1

Varargs and Behavior of Method Arguments

January 27, 2007

Varargs and Behavior of Method Arguments

Varargs

Java 5 brings a convenient new construct to Java that lets a method take zero or more arguments of a given primitive or type. It’s call variable arguments or varargs. In this article we’ll deal with some questions that the introduction of varargs brings up regarding the behavior of method arguments.

A varargs method can be called with zero, one or many individual arguments or with a single array. After the method is called, the arguments are accessed internally using the familiar square brackets array syntax. It looks like this:

public class BasicVarargs {
    public static void main(String[] args) {
        myMethod(); // zero arguments are fine
        myMethod(0); // one argument is fine
        myMethod(0, 1); // multiple arguments are fine
        myMethod(new int[] {0, 1, 2}); // array of args is fine
    }
    static void myMethod(int... myArgs) {
        System.out.println(myArgs[0]);
        System.out.println(myArgs[1]);
        // etc
    }
}

The Issue

This raises a significant question if you’re taking the Sun Certified Java Programmer (SJCP) exam or just want to be knowledgeable about how method parameters are treated. A single argument (whether a primitive or an object reference) cannot be reassigned within a method. Passing a primitive into a method and assigning it a different value will leave the primitive in the calling method unchanged. Passing an object reference into a method and then reassigning the references will, likewise, leave the reference in the calling method unchanged. Objects however, including arrays, have an associated caveat. While the reference to the object cannot be reassigned, the method can use the copy of the reference to alter the content of the object. For a non-array object, this means its instance fields. For an array, this means the primitives or object references at each of its indices. That brings us to our question: If methods with varargs access those arguments as if they were an array, does that mean that they can change the value of a primitive or reassign an object reference that’s held at any given array index? That’s the issue we’re about to explore.

Review of Primitive Argument Behavior

As a quick review, what’s the output of the following?

public class PrimitiveArgBehavior {
	static int myInt = 0;
	public static void main(String[] args) {
	    System.out.println(myInt);
		intMethod(myInt);
		System.out.println(myInt);
	}
	private static void intMethod(int someInt) {
		someInt++;
	}
}

Output:


0
0

myInt remains unchanged. intMethod() works on a copy of myInt and that copy loses scope at the end of intMethod().

Review of Array Reference Argument Behavior

What’s the output of the following?

public class ArrayReferenceArgBehavior {
    static int[] myIntArray = {0};
    public static void main(String[] args) {
        System.out.println(myIntArray[0]);
        intArrayMethod(myIntArray);
        System.out.println(myIntArray[0]);
    }
    static void intArrayMethod(int[] someIntArray) {
        someIntArray = new int[] {1};
    }
}

Output:


0
0

myIntArray remains unchanged. intArrayMethod() works on a copy of the myIntArray reference and that copy loses scope at the end of intArrayMethod(). intArrayMethod() is prevented from reassigning the original reference.

Review of Array Contents Argument Behavior

What’s the output of the following?

public class ArrayContentsArgBehavior {
    static int[] myIntArray = {0};
    public static void main(String[] args) {
        System.out.println(myIntArray[0]);
        intArrayMethod(myIntArray);
        System.out.println(myIntArray[0]);
    }
    static void intArrayMethod(int[] someIntArray) {
        someIntArray[0]++;
    }
}

Output:


0
1

In this case the value did change. While the array reference itself cannot be reassigned to point to another array, the elements of the array can be changed.

Varargs Behavior With Primitives and Arrays

So now we go on to the tricky varargs method. What’s the output of the following?

public class VarArgsBehavior {
	static int myInt = 0;
	static int[] myIntArray = {0};
	public static void main(String[] args) {
		System.out.println("myInt before: " + myInt);
		intVarArgsMethod(myInt);
		System.out.println("myInt after: " + myInt);
		System.out.println("myIntArray before: " + myIntArray[0]);
		intVarArgsMethod(myIntArray);
		System.out.println("myIntArray after: " + myIntArray[0]);
	}
	static void intVarArgsMethod(int... someVarArgsInt) {
		someVarArgsInt[0]++;
	}
}

Now you see why it’s a little unclear? The first call is using an int. The second call is using an array of ints. In either case though, the intvarArgsMethod() is using array syntax to deal with the arguments. So how does it behave when called with individual primitives? Are they treated like individual primitives or like they were placed into an array before the method call?

The output:


myInt before: 0
myInt after: 0
myIntArray before: 0
myIntArray after: 1

Thankfully, the developers that implemented varargs kept the behavior consistent. Even though vararg parameters are accessed in a manner that makes them look like elements of an array, they are still treated in accordance with how they are declared. Kudos to the varargs implementation team. They could very easily have decided to just handle the argument like an array at all times and that would have resulted in some very odd behavior in special cases.

Conclusion

Bottom line, expect varargs to treat an argument in the manner in which it was declared. If a varargs method is called with an array, its arguments will be treated like an array. If called with one or more individual arguments, they will be treated like individual arguments. Varargs respects the manner in which a method is called.

Joshua Smith


References


Sun Guide to Varargs


SCJP Sun Certified Java Programmer Study Guide

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