JAVA vs C++


Contents


In Java, every variable, constant, and function (including main) must be inside some class. Here's a simple example program:

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

Things to note:
  1. There is no final semi-colon at the end of the class definition.

  2. Function main is a member of class Test.

  3. In general, main must:

  4. To write to standard output, you can use either of the following:
    	System.out.println( ... )
    	System.out.print( ... )
        
    The former prints the given expression followed by a newline, while the latter just prints the given expression.

    Like the C++ << operator, these functions can be used to print values of any type. For example, all of the following are fine:

    	System.out.print("hello");   // print a String
    	System.out.print(16);        // print an integer
    	System.out.print(5.5 * .2);  // print a floating-point number
           

    The + operator can be useful when printing. It is overloaded to work on Strings as follows:

    If either operand is a String, it

    1. converts the other operand to a String (if necessary)
    2. creates a new String by concatenating both operands

    Example

    	int x = 20, y = 10;
    	System.out.println("x: " + x + "\ny: " + y);
           
    The output is:
           x: 20
           y: 10
           
    This is because the argument to println is an expression of the form:

    The only operator is +, so the expression is evaluated left-to-right (if there were another operator with higher precedence, the sub-expression involving that operator would be evaluated first). The leftmost sub-expression is:

    One operand (the left one) is a String, so the other operand is converted from an int to a String, and the two strings are concatenated, producing the value:

    Evaluation of the argument to println continues, producing as the final value the String shown above (note that \n means the newline character):

    	x: 20
     	y: 10
           

TEST YOURSELF #1

Assume the following declarations have been made:

    int x = 20, y = 10;
What is printed when each of the following statements is executed?
    System.out.println(x + y);
    System.out.println(x + y + "!");
    System.out.println("printing: " + x + y);
    System.out.println("printing: " + x * y);

solution


C++ Files vs Java Files

A C++ programmer deals with source files, object files, and executable files:

Source Files: .h and .cc (or .cpp or .C)

  created by:  you (the programmer)
  contain   :  C++ source code
  two kinds :
        .h (header files)
	       contain class definitions and function specifications
                       (just headers - no bodies)
               must be included by every file that uses the class / calls the
	               functions
        .cc contain implementations of class member functions and "free" functions,
	            including the main function
	

Object Files: .o

  created by:  the compiler, when called w/ -c flag; for example:
			g++ -c main.cc
	       compiles main.cc creating main.o
  contain   :  object code (not executable)
               source code is compiled, but not linked/loaded
  

Executable Files

  created by:  the compiler (no -c flag)
  contain   :  executable code
	 	      Code is compiled if necessary, then linked and loaded.
		      These are the files that you can actually run, just by typing
		      the name of the file.
  name      :  default = a.out
               any other name is possible via the -o flag; for example:
			g++ main.o -o test
 	       creates an executable named "test"
  

A Java programmer deals with source files and bytecode files (no executable files):

Source Files: .java

  created by   :  you (the programmer)
  contain      :  Java source code (one or more classes per file)
  restrictions :
     (1) each source file can contain at most one public class
     (2) if there is a public class, then the class name and file name must match
     
     Examples
     	If a source file contains the following:
		public class Test { ... }
		class Foo { ... }
		class Bar {... }
	then it must be in a file named Test.java
	
	If a source file contains the following:
		class Test { ... }
		class Foo { ... }
		class Bar {... }
	then it can be in any ".java" file
	
Small digression: End of digression.

Bytecode Files: .class

  created by:  the Java compiler
  contain   :  Java bytecodes, ready to be "executed" -- really interpreted -- by
                               the Java interpreter
  names     :  for each class in a source file (both public and non-public classes),
               the compiler creates one ".class" file, where the file name is the
	       same as the class name

  Example
     If a source file contains the following:
	public class Test { ... }
	class Foo { ... }
	class Bar {... }
     then after compiling you will have three files:
     	Test.class
	Foo.class
	Bar.class               
  

Here's how to compile and run the example "hello world" program given above, assuming that it is in a file named "tst.java":

  1. to compile the source code, type: javac tst.java

  2. to run the interpreter using the name of the class whose main function you want to run, type: java Test

Remember, when compiling a program, you type the full file name, including the ".java" extension; when running a program, you just type the name of the class whose main function you want to run.


TEST YOURSELF #2

Write a complete Java program that uses a loop to sum the numbers from 1 to 10 and prints the result like this:

    The sum is: xxx
Note: Use variable declarations, and a for or while loop with the same syntax as in C++.

Make sure that you are able to compile and execute your program!

solution


Java Types

In Java, there are two "categories" of types: primitive types and reference types:

Notes:

  1. no struct, union, enum, unsigned, typedef
  2. arrays and classes are really pointers!!

C++ Arrays vs Java Arrays


TEST YOURSELF #3

Write a Java function called NonZeros, using the header given below. NonZeros should create and return an array of integers containing all of the non-zero values in its parameter A, in the same order that they occur in A.

    public static int[] NonZeros( int [] A )

Write a complete Java program that includes a main function as well as the NonZeros function. The main function should test NonZeros by creating several arrays, and calling NonZeros with each array. It should print the array it passes to NonZeros as well as the returned array. So for example, when you run your program, your output might look like this (if your NonZeros function is implemented correctly):

    passing [0,1,2,3,2] got back [1,2,3,2]
    passing [0,0] got back []
    passing [22,0,-5,0,126] got back [22,-5,126]
    passing [1,0] got back [1]

solution



TEST YOURSELF #4

For each of the following code fragments, fill in the number of the picture that best illustrates the value of A after the code executes, or fill in "error" to indicate that executing the code causes a runtime error. (In the pictures, a diagonal line indicates a null pointer.)
Code Corresponding Picture or Error
int [] A;
int [] A = new int [4];
int [][] A = new int[4][3];
int [][] A = new int[4][];
A[1] = new int[4];
A[3] = new int[2];
int [] A = new int[4];
int [] B = {0,1,2,3,4,5,6,7,8,9};
System.arraycopy(B,2,A,0,4);
int [] A = new int[4];
int [] B = {2,3,4};
System.arraycopy(B,0,A,0,4);
int [] A = new int[4];
int [] B = {0,1,2,3,4,5,6,7,8,9};
System.arraycopy(B,8,A,0,4);
int [] A = {1,1,1,1};
int [] B = {2,2,2};
System.arraycopy(A,0,B,1,2);
System.arraycopy(B,0,A,0,3);
int [] A = new int[4];
int [] B = {0,1,2,3,4,5,6,7,8,9};
System.arraycopy(B,0,A,0,10);
int [][] A = new int[4][3];
int [] B = {1,2,3,4,5,6,7,8,9,10};
System.arraycopy(B,0,A[0],0,3);
System.arraycopy(B,1,A[1],0,3);
System.arraycopy(B,2,A[2],0,3);
System.arraycopy(B,3,A[3],0,3);

PICTURES

(1) (2) (3)

(4) (5) (6)

(7) (8) (9)

(10) (11) (12)
(13) (14) (15)

solution


C++ Classes vs Java Classes

Aliasing Problems in Java

The fact that arrays and classes are really pointers in Java can lead to some problems:

Problem 1: Simple assignment causes aliasing:

Java code                          conceptual picture
                                          (all empty boxes contain zeros)

                                      +--+       +---+---+---+
int [] A = new int[3],             A: | -|-----> |   |   |   |
                                      +--+       +---+---+---+

       B = new int[2];                +--+       +---+---+
                                   B: | -|-----> |   |   |
                                      +--+       +---+---+

                                      +--+       +---+---+---+
       A[0] = 5;                   A: | -|-----> | 5 |   |   |
                                      +--+       +---+---+---+

                                      +--+       +---+---+---+
       B = A;                      A: | -|-----> | 5 |   |   |
                                      +--+       +---+---+---+
				                 ^
				      +--+       |
				   B: | -|-------+
				      +--+

                                      +--+       +---+---+---+ *** NOTE **
        B[0] = 2;                  A: | -|-----> | 2 |   |   | the value of A[0]
                                      +--+       +---+---+---+ changed, too!
				                 ^
				      +--+       |
				   B: | -|-------+
				      +--+

Problem 2: In Java, all parameters are passed by value, but for arrays and classes the actual parameter is really a pointer, so changing

inside the function does change the actual parameter's element or field. For example:
void f( int [] A )
{
    A[0] = 10;   // change an element of parameter A
    A = null;    // change A itself
}

void g()
{
    int [] B = new int [3];
    B[0] = 5;
    f(B);
    *** B is not null here, because B itself was passed by value
    *** however, B[0] is now 10, because function f changed the first element
    *** of the array
}

In C++, similar problems can arise when a class that has pointer data members is passed by value. This problem is addressed by the use of copy constructors, which can be defined to make copies of the values pointed to, rather than just making copies of the pointers. In Java, the solution is to use the arraycopy operation, or to use a class's clone operation. Cloning will be discussed later.
TEST YOURSELF #5

For each of the following Java code fragments, say whether it causes a compile-time error, a run-time error, or no error. If there is an error, explain why.

   1. int A[5];

   2. int [] A, B;
      B = 0;

   3. int [] A = {1,2,3};
      int [] B;
      B = A;

   4. int [] A;
      A[0] = 0;

   5. int [] A = new int[20];
      int [] B = new int[10];
      A = B;
      A[15] = 0;

solution


Type Conversion

Java is much more limited than C++ in the type conversions that are allowed. Here we discuss conversions among primitive types. Conversions among class objects will be discussed later.

Booleans cannot be converted to other types. For the other primitive types (char, byte, short, int, long, float, and double), there are two kinds of conversion: implicit and explicit.

Implicit conversions: An implicit conversion means that a value of one type is changed to a value of another type without any special directive from the programmer. A char can be implicitly converted to an int, a long, a float, or a double. For example, the following will compile without error:

For the other (numeric) primitive types, the basic rule is that implicit conversions can be done from one type to another if the range of values of the first type is a subset of the range of values of the second type. For example, a byte can be converted to a short, int, long or float; a short can be converted to an int, long, float, or double, etc.

Explicit conversions: Explicit conversions are done via casting: the name of the type to which you want a value converted is given, in parentheses, in front of the value. For example, the following code uses casts to convert a value of type double to a value of type int, and to convert a value of type double to a value of type short:

Casting can be used to convert among any of the primitive types except boolean. Note, however, that casting can lose information; for example, floating-point values are truncated when they are cast to integers (e.g., the value of k in the code fragment given above is 5), and casting among integer types can produce wildly different values (because upper bits, possibly including the sign bit, are lost). So use explicit casting carefully!
TEST YOURSELF #6

Fill in the table below as follows:

The first row has been filled in as an example.

Declaration Correct Rewrite with cast Never correct Variable's value
double d = 5; X 5.0
int k = 5.6;
long x = 5.4;
short n = 99999;
int b = true;
char c = 97;
short s = -10.0;

solution


Solutions to Self-Study Questions

Test Yourself #1

Test Yourself #2

Test Yourself #3

Test Yourself #4

Test Yourself #5

Test Yourself #6