(This pitfall applies equally to all primitive wrapper types, but we will illustrate it for Integer and int.)

When working with Integer objects, it is tempting to use == to compare values, because that is what you would do with int values. And in some cases this will seem to work:

Integer int1_1 = Integer.valueOf("1");
Integer int1_2 = Integer.valueOf(1);

System.out.println("int1_1 == int1_2: " + (int1_1 == int1_2));          // true
System.out.println("int1_1 equals int1_2: " + int1_1.equals(int1_2));   // true

Here we created two Integer objects with the value 1 and compare them (In this case we created one from a String and one from an int literal. There are other alternatives). Also, we observe that the two comparison methods (== and equals) both yield true.

This behavior changes when we choose different values:

Integer int2_1 = Integer.valueOf("1000");
Integer int2_2 = Integer.valueOf(1000);

System.out.println("int2_1 == int2_2: " + (int2_1 == int2_2));          // false
System.out.println("int2_1 equals int2_2: " + int2_1.equals(int2_2));   // true

In this case, only the equals comparison yields the correct result.

The reason for this difference in behavior is, that the JVM maintains a cache of Integer objects for the range -128 to 127. (The upper value can be overridden with the system property "java.lang.Integer.IntegerCache.high" or the JVM argument "-XX:AutoBoxCacheMax=size"). For values in this range, the Integer.valueOf() will return the cached value rather than creating a new one.

Thus, in the first example the Integer.valueOf(1) and Integer.valueOf("1") calls returned the same cached Integer instance. By contrast, in the second example the Integer.valueOf(1000) and Integer.valueOf("1000") both created and returned new Integer objects.

The == operator for reference types tests for reference equality (i.e. the same object). Therefore, in the first example int1_1 == int1_2 is true because the references are the same. In the second example int2_1 == int2_2 is false because the references are different.

A common mistake for Java beginners is to use the == operator to test if two strings are equal. For example:

public class Hello {
    public static void main(String[] args) {
        if (args.length > 0) {
            if (args[0] == "hello") {
                System.out.println("Hello back to you");
            } else {
                System.out.println("Are you feeling grumpy today?");
            }
        }
    }
}

The above program is supposed to test the first command line argument and print different messages when it and isn't the word "hello". But the problem is that it won't work. That program will output "Are you feeling grumpy today?" no matter what the first command line argument is.

In this particular case the String "hello" is put in the string pool while the String args[0] resides on the heap. This means there are two objects representing the same literal, each with its reference. Since == tests for references, not actual equality, the comparison will yield a false most of the times. This doesn't mean that it will always do so.

When you use == to test strings, what you are actually testing is if two String objects are the same Java object. Unfortunately, that is not what string equality means in Java. In fact, the correct way to test strings is to use the equals(Object) method. For a pair of strings, we usually want to test if they consist of the same characters in the same order.

public class Hello2 {
    public static void main(String[] args) {
        if (args.length > 0) {
            if (args[0].equals("hello")) {
                System.out.println("Hello back to you");
            } else {
                System.out.println("Are you feeling grumpy today?");
            }
        }
    }
}

But it actually gets worse. The problem is that == will give the expected answer in some circumstances. For example

public class Test1 {
    public static void main(String[] args) {
        String s1 = "hello";
        String s2 = "hello";
        if (s1 == s2) {
            System.out.println("same");
        } else {
            System.out.println("different");
        }
    }
}

Interestingly, this will print "same", even though we are testing the strings the wrong way. Why is that? Because the Java Language Specification (Section 3.10.5: String Literals) (opens new window) stipulates that any two string >>literals<< consisting of the same characters will actually be represented by the same Java object. Hence, the == test will give true for equal literals. (The string literals are "interned" and added to a shared "string pool" when your code is loaded, but that is actually an implementation detail.)

To add to the confusion, the Java Language Specification also stipulates that when you have a compile-time constant expression that concatenates two string literals, that is equivalent to a single literal. Thus:

   public class Test1 {
    public static void main(String[] args) {
        String s1 = "hello";
        String s2 = "hel" + "lo";
        String s3 = " mum";
        if (s1 == s2) {
            System.out.println("1. same");
        } else {
            System.out.println("1. different");
        }
        if (s1 + s3 == "hello mum") {
            System.out.println("2. same");
        } else {
            System.out.println("2. different");
        }
    }
}

This will output "1. same" and "2. different". In the first case, the + expression is evaluated at compile time and we compare one String object with itself. In the second case, it is evaluated at run time and we compare two different String objects

In summary, using == to test strings in Java is almost always incorrect, but it is not guaranteed to give the wrong answer.

Every time a program opens a resource, such as a file or network connection, it is important to free the resource once you are done using it. Similar caution should be taken if any exception were to be thrown during operations on such resources. One could argue that the FileInputStream (opens new window) has a finalizer (opens new window) that invokes the close() method on a garbage collection event; however, since we can’t be sure when a garbage collection cycle will start, the input stream can consume computer resources for an indefinite period of time. The resource must be closed in a finally section of a try-catch block:

private static void printFileJava6() throws IOException {
    FileInputStream input;
    try {
        input = new FileInputStream("file.txt");
        int data = input.read();
        while (data != -1){
            System.out.print((char) data);
            data = input.read();
        }
    } finally {
        if (input != null) {
            input.close();
        }
    }
}

Since Java 7 there is a really useful and neat statement introduced in Java 7 particularly for this case, called try-with-resources:

private static void printFileJava7() throws IOException {
    try (FileInputStream input = new FileInputStream("file.txt")) {
        int data = input.read();
        while (data != -1){
            System.out.print((char) data);
            data = input.read();
        }
    }
}

The try-with-resources statement can be used with any object that implements the Closeable or AutoCloseable interface. It ensures that each resource is closed by the end of the statement. The difference between the two interfaces is, that the close() method of Closeable throws an IOException which has to be handled in some way.

In cases where the resource has already been opened but should be safely closed after use, one can assign it to a local variable inside the try-with-resources

private static void printFileJava7(InputStream extResource) throws IOException {
    try (InputStream input = extResource) {
        ... //access resource
    }
}

The local resource variable created in the try-with-resources constructor is effectively final.

Some people recommend that you should apply various tests to a file before attempting to open it either to provide better diagnostics or avoid dealing with exceptions. For example, this method attempts to check if path corresponds to a readable file:

public static File getValidatedFile(String path) throws IOException {
    File f = new File(path);
    if (!f.exists()) throw new IOException("Error: not found: " + path);
    if (!f.isFile()) throw new IOException("Error: Is a directory: " + path);
    if (!f.canRead()) throw new IOException("Error: cannot read file: " + path);
    return f;
}

You might use the above method like this:

File f = null;
try {
    f = getValidatedFile("somefile");
} catch (IOException ex) {
    System.err.println(ex.getMessage());
    return;
}
try (InputStream is = new FileInputStream(file)) {
    // Read data etc.
}

The first problem is in the signature for FileInputStream(File) because the compiler will still insist we catch IOException here, or further up the stack.

The second problem is that checks performed by getValidatedFile do not guarantee that the FileInputStream will succeed.

The third problem is that the tests are inefficient. For example, the exists, isFile and canRead calls will each make a syscall (opens new window) to perform the required check. Another syscall is then made to open the file, which repeats the same checks behind the scenes.

In short, methods like getValidatedFile are misguided. It is better to simply attempt to open the file and handle the exception:

try (InputStream is = new FileInputStream("somefile")) {
    // Read data etc.
} catch (IOException ex) {
    System.err.println("IO Error processing 'somefile': " + ex.getMessage());
    return;
}

If you wanted to distinguish IO errors thrown while opening and reading, you could use a nested try / catch. If you wanted to produce better diagnostics for open failures, you could perform the exists, isFile and canRead checks in the handler.

No Java variable represents an object.

String foo;   // NOT AN OBJECT

Neither does any Java array contain objects.

String bar[] = new String[100];  // No member is an object.

If you mistakenly think of variables as objects, the actual behavior of the Java language will surprise you.

Variables are not objects in either case, and they don't contain objects in either case. They may contain references to objects, but that is saying something different.

Example class

The examples that follow use this class, which represents a point in 2D space.

public final class MutableLocation {
   public int x;
   public int y;

   public MutableLocation(int x, int y) {
       this.x = x;
       this.y = y;
   }

   public boolean equals(Object other) {
       if (!(other instanceof MutableLocation) {
           return false;
       }
       MutableLocation that = (MutableLocation) other;
       return this.x == that.x && this.y == that.y;
   }
}

An instance of this class is an object that has two fields x and y which have the type int.

We can have many instances of the MutableLocation class. Some will represent the same locations in 2D space; i.e. the respective values of x and y will match. Others will represent different locations.

Multiple variables can point to the same object

MutableLocation here = new MutableLocation(1, 2);
 MutableLocation there = here;
 MutableLocation elsewhere = new MutableLocation(1, 2);

In the above, we have declared three variables here, there and elsewhere that can hold references to MutableLocation objects.

If you (incorrectly) think of these variables as being objects, then you are likely to misread the statements as saying:

- Copy the location "[1, 2]" to `here` - Copy the location "[1, 2]" to `there` - Copy the location "[1, 2]" to `elsewhere`

From that, you are likely to infer we have three independent objects in the three variables. In fact there are only two objects created by the above. The variables here and there actually refer to the same object.

We can demonstrate this. Assuming the variable declarations as above:

System.out.println("BEFORE: here.x is " + here.x + ", there.x is " + there.x +
                   "elsewhere.x is " + elsewhere.x);
here.x = 42;
System.out.println("AFTER: here.x is " + here.x + ", there.x is " + there.x +
                   "elsewhere.x is " + elsewhere.x);

This will output the following:

BEFORE: here.x is 1, there.x is 1, elsewhere.x is 1
AFTER: here.x is 42, there.x is 42, elsewhere.x is 1

We assigned a new value to here.x and it changed the value that we see via there.x. They are referring to the same object. But the value that we see via elsewhere.x has not changed, so elsewhere must refer to a different object.

If a variable was an object, then the assignment here.x = 42 would not change there.x.

The equality operator does NOT test that two objects are equal

Applying the equality (==) operator to reference values tests if the values refer to the same object. It does not test whether two (different) objects are "equal" in the intuitive sense.

MutableLocation here = new MutableLocation(1, 2);
 MutableLocation there = here;
 MutableLocation elsewhere = new MutableLocation(1, 2);

 if (here == there) {
     System.out.println("here is there");
 }
 if (here == elsewhere) {
     System.out.println("here is elsewhere");
 }

This will print "here is there", but it won't print "here is elsewhere". (The references in here and elsewhere are for two distinct objects.)

By contrast, if we call the equals(Object) method that we implemented above, we are going to test if two MutableLocation instances have an equal location.

if (here.equals(there)) {
     System.out.println("here equals there");
 }
 if (here.equals(elsewhere)) {
     System.out.println("here equals elsewhere");
 }

This will print both messages. In particular, here.equals(elsewhere) returns true because the semantic criteria we chose for equality of two MutableLocation objects has been satisfied.

Method calls do NOT pass objects at all

Java method calls use pass by value¹ to pass arguments and return a result.

When you pass a reference value to a method, you're actually passing a reference to an object by value, which means that it is creating a copy of the object reference.

As long as both object references are still pointing to the same object, you can modify that object from either reference, and this is what causes confusion for some.

However, you are not passing an object by reference². The distinction is that if the object reference copy is modified to point to another object, the original object reference will still point to the original object.

void f(MutableLocation foo) {  
    foo = new MutableLocation(3, 4);   // Point local foo at a different object.
}

void g() {
    MutableLocation foo = MutableLocation(1, 2);
    f(foo);
    System.out.println("foo.x is " + foo.x); // Prints "foo.x is 1".
}

Neither are you passing a copy of the object.

void f(MutableLocation foo) {  
    foo.x = 42;
}

void g() {
    MutableLocation foo = new MutableLocation(0, 0);
    f(foo);
    System.out.println("foo.x is " + foo.x); // Prints "foo.x is 42"
}

^{1 - In languages like Python and Ruby, the term "pass by sharing" is preferred for "pass by value" of an object / reference.}

^{2 - The term "pass by reference" or "call by reference" has a very specific meaning in programming language terminology. In effect, it means that you pass the address of a variable or an array element, so that when the called method assigns a new value to the formal argument, it changes the value in the original variable. Java does not support this. For a more fulsome description of different mechanisms for passing parameters, please refer to https://en.wikipedia.org/wiki/Evaluation_strategy (opens new window).}

Java manages memory automatically. You are not required to free memory manually. An object's memory on the heap may be freed by a garbage collector when the object is no longer reachable by a live thread.

However, you can prevent memory from being freed, by allowing objects to be reachable that are no longer needed. Whether you call this a memory leak or memory packratting, the result is the same -- an unnecessary increase in allocated memory.

Memory leaks in Java can happen in various ways, but the most common reason is everlasting object references, because the garbage collector can’t remove objects from the heap while there are still references to them.

Static fields

One can create such a reference by defining class with a static field containing some collection of objects, and forgetting to set that static field to null after the collection is no longer needed. static fields are considered GC roots and are never collected. Another issue is leaks in non-heap memory when JNI (opens new window) is used.

Classloader leak

By far, though, the most insidious type of memory leak is the classloader leak (opens new window). A classloader holds a reference to every class it has loaded, and every class holds a reference to its classloader. Every object holds a reference to its class as well. Therefore, if even a single object of a class loaded by a classloader is not garbage, not a single class that that classloader has loaded can be collected. Since each class also refers to its static fields, they cannot be collected either.

Accumulation leak The accumulation leak example could look like the following:

final ScheduledExecutorService scheduledExecutorService = Executors.newScheduledThreadPool(1);
final Deque<BigDecimal> numbers = new LinkedBlockingDeque<>();
final BigDecimal divisor = new BigDecimal(51);

scheduledExecutorService.scheduleAtFixedRate(() -> {
    BigDecimal number = numbers.peekLast();
    if (number != null && number.remainder(divisor).byteValue() == 0) {
        System.out.println("Number: " + number);
        System.out.println("Deque size: " + numbers.size());
    }
}, 10, 10, TimeUnit.MILLISECONDS);

scheduledExecutorService.scheduleAtFixedRate(() -> {
    numbers.add(new BigDecimal(System.currentTimeMillis()));
}, 10, 10, TimeUnit.MILLISECONDS);

try {
    scheduledExecutorService.awaitTermination(1, TimeUnit.DAYS);
} catch (InterruptedException e) {
    e.printStackTrace();
}

This example creates two scheduled tasks. The first task takes the last number from a deque called numbers, and, if the number is divisible by 51, it prints the number and the deque's size. The second task puts numbers into the deque. Both tasks are scheduled at a fixed rate, and they run every 10 ms.

If the code is executed, you’ll see that the size of the deque is permanently increasing. This will eventually cause the deque to be filled with objects that consume all available heap memory.

To prevent this while preserving the semantics of this program, we can use a different method for taking numbers from the deque: pollLast. Contrary to the method peekLast, pollLast returns the element and removes it from the deque while peekLast only returns the last element.

New Java programmers often forget, or fail to fully comprehend, that the Java String class is immutable. This leads to problems like the one in the following example:

public class Shout {
    public static void main(String[] args) {
        for (String s : args) {
            s.toUpperCase();
            System.out.print(s);
            System.out.print(" ");
        }
        System.out.println();
    }
}

The above code is supposed to print command line arguments in upper case. Unfortunately, it does not work, the case of the arguments is not changed. The problem is this statement:

s.toUpperCase();

You might think that calling toUpperCase() is going to change s to an uppercase string. It doesn't. It can't! String objects are immutable. They cannot be changed.

In reality, the toUpperCase() method returns a String object which is an uppercase version of the String that you call it on. This will probably be a new String object, but if s was already all uppercase, the result could be the existing string.

So in order to use this method effectively, you need to use the object returned by the method call; for example:

s = s.toUpperCase();

In fact, the "strings never change" rule applies to all String methods. If you remember that, then you can avoid a whole category of beginner's mistakes.

Occasionally we see StackOverflow Java questions (and C or C++ questions) that ask what something like this:

i += a[i++] + b[i--];

evaluates to ... for some known initial states of i, a and b.

Generally speaking:

• for Java the answer is always specified¹, but non-obvious, and often difficult to figure out
• for C and C++ the answer is often unspecified.

Such examples are often used in exams or job interviews as an attempt to see if the student or interviewee understands how expression evaluation really works in the Java programming language. This is arguably legitimate as a "test of knowledge", but that does not mean that you should ever do this in a real program.

To illustrate, the following seemingly simple example has appeared a few times in StackOverflow questions (like this one (opens new window)). In some cases, it appears as a genuine mistake in someone's code.

int a = 1;
a = a++;
System.out.println(a);    // What does this print.

Most programmers (including Java experts) reading those statements quickly would say that it outputs 2. In fact, it outputs 1. For a detailed explanation of why, please read this Answer (opens new window).

However the real takeaway from this and similar examples is that any Java statement that both assigns to and side-effects the same variable is going to be at best hard to understand, and at worst downright misleading. You should avoid writing code like this.

^{1 - modulo potential issues with the Java Memory Model (opens new window) if the variables or objects are visible to other threads.}