# Java Pitfalls - Language syntax

Several Java programming language misusage might conduct a program to generate incorrect results despite being compiled correctly. This topic main purpose is to list common pitfalls with their causes, and to propose the correct way to avoid falling in such problems.

# Pitfall - Missing a ‘break’ in a 'switch' case

These Java issues can be very embarrassing, and sometimes remain undiscovered until run in production. Fallthrough behavior in switch statements is often useful; however, missing a “break” keyword when such behavior is not desired can lead to disastrous results. If you have forgotten to put a “break” in “case 0” in the code example below, the program will write “Zero” followed by “One”, since the control flow inside here will go through the entire “switch” statement until it reaches a “break”. For example:

public static void switchCasePrimer() {
        int caseIndex = 0;
        switch (caseIndex) {
            case 0:
                System.out.println("Zero");
            case 1:
                System.out.println("One");
                break;
            case 2:
                System.out.println("Two");
                break;
            default:
                System.out.println("Default");
        }
}

In most cases, the cleaner solution would be to use interfaces and move code with specific behaviour into separate implementations (composition over inheritance)

If a switch-statement is unavoidable it is recommended to document "expected" fallthroughs if they occur. That way you show fellow developers that you are aware of the missing break, and that this is expected behaviour.

switch(caseIndex) {
    [...]
    case 2:
        System.out.println("Two");
        // fallthrough
    default:
        System.out.println("Default");

# Pitfall - Leaving out braces: the "dangling if" and "dangling else" problems

The latest version of the Oracle Java style guide mandates that the "then" and "else" statements in an if statement should always be enclosed in "braces" or "curly brackets". Similar rules apply to the bodies of various loop statements.

if (a) {           // <- open brace
    doSomething();
    doSomeMore();
}                  // <- close brace

This is not actually required by Java language syntax. Indeed, if the "then" part of an if statement is a single statement, it is legal to leave out the braces

if (a)
    doSomething();

or even

if (a) doSomething();

However there are dangers in ignoring Java style rules and leaving out the braces. Specifically, you significantly increase the risk that code with faulty indentation will be misread.

The "dangling if" problem:

Consider the example code from above, rewritten without braces.

if (a)
   doSomething();
   doSomeMore();

This code seems to say that the calls to doSomething and doSomeMore will both occur if and only if a is true. In fact, the code is incorrectly indented. The Java Language Specification that the doSomeMore() call is a separate statement following the if statement. The correct indentation is as follows:

if (a)
   doSomething();
doSomeMore();

The "dangling else" problem

A second problem appears when we add else to the mix. Consider the following example with missing braces.

if (a)
   if (b)
      doX();
   else if (c)
      doY(); 
else
   doZ();

The code above seems to say that doZ will be called when a is false. In fact, the indentation is incorrect once again. The correct indentation for the code is:

if (a)
   if (b)
      doX();
   else if (c)
      doY(); 
   else
      doZ();

If the code was written according to the Java style rules, it would actually look like this:

if (a) {
   if (b) {
      doX();
   } else if (c) {
      doY(); 
   } else {
      doZ();
   }
}

To illustrate why that is better, suppose that you had accidentally mis-indented the code. You might end up with something like this:

if (a) {                         if (a) {
   if (b) {                          if (b) {
      doX();                            doX();
   } else if (c) {                   } else if (c) {
      doY();                            doY();
} else {                         } else {
   doZ();                            doZ();
}                                    }
}                                }

But in both cases, the mis-indented code "looks wrong" to the eye of an experienced Java programmer.

# Pitfall - Declaring classes with the same names as standard classes

Sometimes, programmers who are new to Java make the mistake of defining a class with a name that is the same as a widely used class. For example:

package com.example;

/**
 * My string utilities
 */
public class String {
    ....
}

Then they wonder why they get unexpected errors. For example:

package com.example;

public class Test {
    public static void main(String[] args) {
        System.out.println("Hello world!");
    }
}

If you compile and then attempt to run the above classes you will get an error:

$ javac com/example/*.java
$ java com.example.Test
Error: Main method not found in class test.Test, please define the main method as:
   public static void main(String[] args)
or a JavaFX application class must extend javafx.application.Application

Someone looking at the code for the Test class would see the declaration of main and look at its signature and wonder what the java command is complaining about. But in fact, the java command is telling the truth.

When we declare a version of String in the same package as Test, this version takes precedence over the automatic import of java.lang.String. Thus, the signature of the Test.main method is actually

void main(com.example.String[] args) 

instead of

void main(java.lang.String[] args)

and the java command will not recognize that as an entrypoint method.

Lesson: Do not define classes that have the same name as existing classes in java.lang, or other commonly used classes in the Java SE library. If you do that, you are setting yourself open for all sorts of obscure errors.

# Pitfall - Ignoring method visibility

Even experienced Java developers tend to think that Java has only three protection modifiers. The language actually has four! The package private (a.k.a. default) level of visibility is often forgotten.

You should pay attention to what methods you make public. The public methods in an application are the application’s visible API. This should be as small and compact as possible, especially if you are writing a reusable library (see also the SOLID (opens new window) principle). It is important to similarly consider the visibility of all methods, and to only use protected or package private access where appropriate.

When you declare methods that should be private as public, you expose the internal implementation details of the class.

A corollary to this is that you only unit test (opens new window) the public methods of your class - in fact you can only test public methods. It is bad practice to increase the visibility of private methods just to be able to run unit tests against those methods. Testing public methods that call the methods with more restrictive visibility should be sufficient to test an entire API. You should never expand your API with more public methods only to allow unit testing.

# Pitfall - Octal literals

Consider the following code snippet:

// Print the sum of the numbers 1 to 10
int count = 0;
for (int i = 1; i < 010; i++) {    // Mistake here ....
    count = count + i;
}
System.out.println("The sum of 1 to 10 is " + count);

A Java beginner might be surprised to know that the above program prints the wrong answer. It actually prints the sum of the numbers 1 to 8.

The reason is that an integer literal that starts with the digit zero ('0') is interpreted by the Java compiler as an octal literal, not a decimal literal as you might expect. Thus, 010 is the octal number 10, which is 8 in decimal.

# Pitfall - Using '==' to test a boolean

Sometimes a new Java programmer will write code like this:

public void check(boolean ok) {
    if (ok == true) {           // Note 'ok == true'
        System.out.println("It is OK");
    }
}

An experienced programmer would spot that as being clumsy and want to rewrite it as:

public void check(boolean ok) {
    if (ok) {
       System.out.println("It is OK");
    }
}

However, there is more wrong with ok == true than simple clumsiness. Consider this variation:

public void check(boolean ok) {
    if (ok = true) {           // Oooops!
        System.out.println("It is OK");
    }
}

Here the programmer has mistyped == as = ... and now the code has a subtle bug. The expression x = true unconditionally assigns true to x and then evaluates to true. In other words, the check method will now print "It is OK" no matter what the parameter was.

The lesson here is to get out of the habit of using == false and == true. In addition to being verbose, they make your coding more error prone.

Note: A possible alternative to ok == true that avoids the pitfall is to use Yoda conditions (opens new window); i.e. put the literal on the left side of the relational operator, as in true == ok. This works, but most programmers would probably agree that Yoda conditions look odd. Certainly ok (or !ok) is more concise and more natural.

# Pitfall - Misplaced semicolons and missing braces

This is a mistake that causes real confusion for Java beginners, at least the first time that they do it. Instead of writing this:

if (feeling == HAPPY)
    System.out.println("Smile");
else
    System.out.println("Frown");

they accidentally write this:

if (feeling == HAPPY);
    System.out.println("Smile");
else
    System.out.println("Frown");

and are puzzled when the Java compiler tells them that the else is misplaced. The Java compiler with interpret the above as follows:

if (feeling == HAPPY)
    /*empty statement*/ ;
System.out.println("Smile");   // This is unconditional
else                           // This is misplaced.  A statement cannot
                               // start with 'else'
System.out.println("Frown");

In other cases, there will be no be compilation errors, but the code won't do what the programmer intends. For example:

for (int i = 0; i < 5; i++);
    System.out.println("Hello");

only prints "Hello" once. Once again, the spurious semicolon means that the body of the for loop is an empty statement. That means that the println call that follows is unconditional.

Another variation:

for (int i = 0; i < 5; i++);
    System.out.println("The number is " + i);

This will give a "Cannot find symbol" error for i. The presence of the spurious semicolon means that the println call is attempting to use i outside of its scope.

In those examples, there is a straight-forward solution: simply delete the spurious semicolon. However, there are some deeper lessons to be drawn from these examples:

  • The semicolon in Java is not "syntactic noise". The presence or absence of a semicolon can change the meaning of your program. Don't just add them at the end of every line.
  • Don't trust your code's indentation. In the Java language, extra whitespace at the beginning of a line is ignored by the compiler.
  • Use an automatic indenter. All IDEs and many simple text editors understand how to correctly indent Java code.
  • This is the most important lesson. Follow the latest Java style guidelines, and put braces around the "then" and "else" statements and the body statement of a loop. The open brace (`{`) should not be on a new line.
  • If the programmer followed the style rules then the if example with a misplaced semicolons would look like this:

    if (feeling == HAPPY); {
        System.out.println("Smile");
    } else {
        System.out.println("Frown");
    }
    
    

    That looks odd to an experienced eye. If you auto-indented that code, it would probably look like this:

    if (feeling == HAPPY); {
                               System.out.println("Smile");
                           } else {
                               System.out.println("Frown");
                           }
    
    

    which should stand out as wrong to even a beginner.

    # Pitfall - Wildcard imports can make your code fragile

    Consider the following partial example:

    import com.example.somelib.*;
    import com.acme.otherlib.*;
    
    public class Test {
        private Context x = new Context();   // from com.example.somelib
        ...
    }
    
    

    Suppose that when when you first developed the code against version 1.0 of somelib and version 1.0 of otherlib. Then at some later point, you need to upgrade your dependencies to a later versions, and you decide to use otherlib version 2.0. Also suppose that one of the changes that they made to otherlib between 1.0 and 2.0 was to add a Context class.

    Now when you recompile Test, you will get a compilation error telling you that Context is an ambiguous import.

    If you are familiar with the codebase, this probably is just a minor inconvenience. If not, then you have some work to do to address this problem, here and potentially elsewhere.

    The problem here is the wildcard imports. On the one hand, using wildcards can make your classes a few lines shorter. On the other hand:

  • Upwards compatible changes to other parts of your codebase, to Java standard libraries or to 3rd party libraries can lead to compilation errors.
  • Readability suffers. Unless you are using an IDE, figuring out which of the wildcard imports is pulling in a named class can be difficult.
  • The lesson is that it is a bad idea to use wildcard imports in code that needs to be long lived. Specific (non-wildcard) imports are not much effort to maintain if you use an IDE, and the effort is worthwhile.

    # Pitfall: Using 'assert' for argument or user input validation

    A question that occasionally on StackOverflow is whether it is appropriate to use assert to validate arguments supplied to a method, or even inputs provided by the user.

    The simple answer is that it is not appropriate.

    Better alternatives include:

    • Throwing an IllegalArgumentException using custom code.
    • Using the Preconditions methods available in Google Guava library.
    • Using the Validate methods available in Apache Commons Lang3 library.

    This is what the Java Language Specification (JLS 14.10, for Java 8) (opens new window) advises on this matter:

    Typically, assertion checking is enabled during program development and testing, and disabled for deployment, to improve performance. Because assertions may be disabled, programs must not assume that the expressions contained in assertions will be evaluated. Thus, these boolean expressions should generally be free of side effects. Evaluating such a boolean expression should not affect any state that is visible after the evaluation is complete. It is not illegal for a boolean expression contained in an assertion to have a side effect, but it is generally inappropriate, as it could cause program behavior to vary depending on whether assertions were enabled or disabled. In light of this, assertions should not be used for argument checking in public methods. Argument checking is typically part of the contract of a method, and this contract must be upheld whether assertions are enabled or disabled. A secondary problem with using assertions for argument checking is that erroneous arguments should result in an appropriate run-time exception (such as IllegalArgumentException, ArrayIndexOutOfBoundsException, or NullPointerException). An assertion failure will not throw an appropriate exception. Again, it is not illegal to use assertions for argument checking on public methods, but it is generally inappropriate. It is intended that AssertionError never be caught, but it is possible to do so, thus the rules for try statements should treat assertions appearing in a try block similarly to the current treatment of throw statements.

    # Pitfall - Overloading instead of overriding

    Consider the following example:

    public final class Person {
        private final String firstName;
        private final String lastName;
       
        public Person(String firstName, String lastName) {
            this.firstName = (firstName == null) ? "" : firstName;
            this.lastName = (lastName == null) ? "" : lastName;
        }
    
        public boolean equals(String other) {
            if (!(other instanceof Person)) {
                return false;
            }
            Person p = (Person) other;
            return firstName.equals(p.firstName) &&
                    lastName.equals(p.lastName);
        }
    
        public int hashcode() {
            return firstName.hashCode() + 31 * lastName.hashCode();
        }
    }
    
    

    This code is not going to behave as expected. The problem is that the equals and hashcode methods for Person do not override the standard methods defined by Object.

    • The equals method has the wrong signature. It should be declared as equals(Object) not equals(String).
    • The hashcode method has the wrong name. It should be hashCode() (note the capital C).

    These mistakes mean that we have declared accidental overloads, and these won't be used if Person is used in a polymorphic context.

    However, there is a simple way to deal with this (from Java 5 onwards). Use the @Override annotation whenever you intend your method to be an override:

    public final class Person {
        ...
    
        @Override
        public boolean equals(String other) {
            ....
        }
    
        @Override
        public hashcode() {
            ....
        }
    }
    
    

    When we add an @Override annotation to a method declaration, the compiler will check that the method does override (or implement) a method declared in a superclass or interface. So in the example above, the compiler will give us two compilation errors, which should be enough to alert us to the mistake.

    # Pitfall of Auto-Unboxing Null Objects into Primitives

    public class Foobar {
        public static void main(String[] args) {
    
            // example: 
            Boolean ignore = null;
            if (ignore == false) {
                System.out.println("Do not ignore!");
            }
        }
    }
    
    

    The pitfall here is that null is compared to false. Since we're comparing a primitive boolean against a Boolean, Java attempts to unbox the the Boolean Object into a primitive equivalent, ready for comparison. However, since that value is null, a NullPointerException is thrown.

    Java is incapable of comparing primitive types against null values, which causes a NullPointerException at runtime. Consider the primitive case of the condition false == null; this would generate a compile time error incomparable types: int and <null>.

    # Remarks

    This topic is about specific aspects of the Java language syntax that are either error prone or that that should not be used in certain ways.