CodeGym /Java Course /Module 1. Java Syntax /Method parameters in Java

Method parameters in Java

Module 1. Java Syntax
Level 9 , Lesson 1
Available

1. Passing arguments

And now the fun begins. You probably already know from methods like System.out.println() that we can pass arguments to methods. Once we are inside the method, we refer to them as parameters. In fact, parameters greatly enhance the benefits we get from creating and using methods.

How do we declare a method with parameters? It's actually quite simple:

public static void name(parameters)
{
  method body
}

Where name is the unique name of the method and method body represents the commands that make up the method. And parameters is a placeholder for the method parameters, separated by commas. Let's describe this template in more detail:

public static void name(Type1 name1, Type2 name2, Type3 name3)
{
  method body
}

Examples:

Code Explanation
public static void print(String str)
{
}
The print method is declared with a parameter:
String str
public static void print(String str, int count)
{
}
The print method is declared with two parameters:
String str
int count
public static void write(int x, int y)
{
}
The write method is declared with two parameters:
int x
int y

If we don't want the method to have parameters, then we just leave the parentheses empty.

Parameters are special variables within a method. With their help, you can pass various values to the method when it is called.

For example, let's write a method that displays a string of text a given number of times.

You already know how to write code to display a string on the screen several times. But which string should you print? And how many times? That's what we need the parameters for.

The code that does this would look like this:

Code Explanation
class Solution
{
   public static void printLines(String text, int count)
   {
     for (int i = 0; i < count; i++)
       System.out.println(text);
   }

   public static void main(String[] args)
   {
     printLines("Hi", 10);
     printLines("Bye", 20);
   }
}


We declared the printLines method with the following parameters:
String text, int count
The method displays String text count times





We call the printLines method with various parameters

Each time a method is called, its parameters are assigned the passed values, and only then do we start to execute the commands inside the method.


2. Arguments

I would like to draw your attention a little more to calling a method with parameters.

The values passed to the method are usually called arguments when they are passed to the method.

Let's look at another example:

Code Explanation
class Solution
{
   public static void printLines(String text, int count)
   {
     for (int i = 0; i < count; i++)
       System.out.println(text);
   }

   public static void main(String[] args)
   {
     printLines("Hi", 10);
     printLines("Bye", 20);
   }
}


We declared the printLines method with the following parameters:
String text, int count
The method displays String text count times




We call the printLines method with the following arguments:
text = "Hi"; count = 10;
text = "Bye"; count = 20;

When the printLines method was called for the first time, its parameters were assigned the following values:
String text = "Hi", int count = 10.

When the printLines method was called the second time, its parameters were assigned different values:
String text = "Bye", int count = 20.

Parameters are no more and no less than variables that are assigned certain values when a method is called. The values "Hi", "Bye", 10, and 20 are themselves called arguments."


3. Conflicting variable names when calling a method

Variables can be used as method arguments. This is simple and understandable, but it can potentially produce some difficulties. Let's go back to that same example, but this time we'll move the arguments into separate variables:

Code Explanation
class Solution
{
   public static void printLines(String text, int count)
   {
     for (int i = 0; i < count; i++)
       System.out.print(text);
   }

   public static void main(String[] args)
   {
     String str = "Hi";
     int n = 10;
     printLines(str, n);
   }
}


We declared the printLines method with the following parameters:
String text, int count
The method displays String text count times







We call the printLines method with the following arguments:
text = str;
count = n;

So far, so good: we have a str variable. Its value is assigned to the text parameter when the method is called. We have an n variable. Its value is assigned to the count parameter when the method is called." So far, everything is clear.

Now let's rename our variables in the main method:

Code Explanation
class Solution
{
   public static void printLines(String text, int count)
   {
     for (int i = 0; i < count; i++)
       System.out.print(text);
   }

   public static void main(String[] args)
   {
     String text = "Hi";
     int count = 10;
     printLines(text, count);
   }
}


We declared the printLines method with the following parameters:
String text, int count
The method displays String text count times







We call the printLines method with the following arguments:
text = text;
count = count;

Pay attention to two things

First: we have variables with the same name in different methods. These are different variables (we deliberately depict them using different colors). Everything works the same as in the previous example, where the variables in the main method were named str and n.

Second: Nothing magical happens when the method is called. The parameters are simply assigned the argument values. Regardless of whether they are numbers, strings, variables, or expressions.

After we rename the variables in the main method, nothing has changed. They were different variables in different methods previously, and so they remain. There is no magic connection between the text and text variables.


9
Task
New Java Syntax, level 9, lesson 1
Locked
Integer literals
Four public fields, corresponding to the four integer types, are declared in the Solution class. When declared, these fields are initialized with various values stored in integer literals. But the program doesn't compile, and you need to fix it. To do this, make the fewest possible changes to the va
9
Task
New Java Syntax, level 9, lesson 1
Locked
Floating point literals
Seven public fields are declared and initialized in the Solution class. They are initialized with various values stored in floating point literals. But the program doesn't compile, and you need to fix it. To do this, change the field types so they match the values. Do not change the field names or v

4. Passing references to methods

I hope you understood everything from the previous lesson, because now we're going to revisit passing arguments to methods, only we'll dive deeper.

As you already know, some Java variables store not the values themselves, but instead a reference, i.e. the address of the block of memory where the values are located. This is how string variables and array variables work.

When you assign another array variable to an array variable, what happens? That's right. The two variables start to refer to the same space in memory:

Passing references to methods

And what happens if one of these variables is a method parameter?

Code Explanation
class Solution
{
   public static void printArraySum(int[] data)
   {
     int sum = 0;
     for (int i = 0; i < data.length; i++)
       sum = sum + data[i];
     System.out.println(sum);
   }
   
   public static void main(String[] args)
   {
     int[] months = {31, 28, 31, 30, 31, 30, 31, 31, 30};
     printArraySum(months);
   }
}


The printArraySum method calculates the sum of the numbers in the passed array and displays it on the screen

Exactly the same thing happens: the data parameter will contain a reference to the same area of memory as the months variable. When the method is called, a simple assignment occurs: data = months.

And since both variables refer to the same area of memory storing an integer, then the printArraySum method can not only read values from the array, but also change them!

For example, we can write our own method that fills a two-dimensional array with the same value. This is how it might look:

Code Explanation
class Solution
{
   public static void fill(int[][] data, int value)
   {
     for (int i = 0; i < data.length; i++)
     {
       for (int j = 0; j < data[i].length; j++)
         data[i][j] = value;
     }
  }

   public static void main(String[] args)
   {
     int[][] months = {{31, 28}, {31, 30, 31}, {30, 31, 31}};
     fill (months, 8);
   }
}


The fill method iterates over every cell in the passed two-dimensional array and assigns value to them.








We create a two-dimensional array.
We fill the entire array with the number 8.

9
Task
New Java Syntax, level 9, lesson 1
Locked
String literals
A public string field is declared and initialized in the Solution class. But the string is too long and difficult to read. For better readability, you need to split it into 5 substrings and concatenate them with the "+" (string concatenation) operator, like this: - first line: "Always write code as

5. Methods with the same name

Now let's return to method names once again.

Java standards require all methods in the same class to have unique names. In other words, it's impossible to declare two identically named methods in the same class.

When methods are compared for sameness, not only are the names taken into account, but also the types of the parameters! Note that the names of the parameters are not taken into account. Examples:

Code Explanation
void fill(int[] data, int value) {
}
void fill(int[][] data, int value) {
}
void fill(int[][][] data, int value)  {
}
These three methods are different methods. They can be declared in the same class.
void print(String str) {
}
void print(String str, String str2) {
}
void print(int val) {
}
void print(double val) {
}
void print() {
}
Each of these five methods is considered different. They can be declared in the same class.
void sum(int x, int y) {
}
void sum(int data, int value) {
}
These two methods are considered the same, meaning they cannot be declared in the same class.

Why are some methods considered the same, while others are different? And why are parameter names not taken into account when determining the uniqueness of a method?

Why is uniqueness necessary at all? The thing is that when the compiler compiles a program, it must know exactly which method you intend to call in any given place.

For example, if you write System.out.println("Hi"), the compiler is smart and will easily conclude that you intend to call the println() method here with a String parameter.

But if you write System.out.println(1.0), the compiler will see a call to the println() method with a double parameter.

When a method is called, the compiler ensures that the types of the arguments match the types of the parameters. It does not pay any attention to the name of the arguments. In Java, parameter names do not help the compiler determine which method to call. And that's why they are not taken into account when determining the uniqueness of a method.

The name of a method and the types of its parameters are called the method signature. For example, sum(int, int)

Each class must have methods with unique signatures rather than methods with unique names.


9
Task
New Java Syntax, level 9, lesson 1
Locked
Character literals
Four public character fields are declared in the Solution class. Some values have been assigned to them. The program doesn't compile. You need to fix this without changing the character values. All the fields are static — this is necessary in order to access them in the main() method. You can see th
Comments (9)
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haalk3n Level 6, Romania
29 February 2024
Here, have this, they didn't teach you anything about reversing arrays or how to even go about it, but don't break your brain for hours when you can look at this, understand it, then move on with your life. Garbage exercise. //write your code here int i, j, len; // j is placeholder for a number len = array.length; for (i = 0; i < len / 2; i++) { j = array[i]; array[i] = array[len - i - 1]; array[len - i - 1] = j; }
Erik Flo Level 10, Stockholm, Sweden
10 April 2024
Exactly the same code you just create variables so you can skip writing array.length multiple times. This is just like calling a person an idiot cause they use a synonym of a word you are used to use...
Abe Level 13, los angeles, United States
21 January 2023
same confusion here, I did the same thing as Ziedins and was wondering what I was doing wrong...
Edgars Ziedins Level 8, Latvia
25 November 2022
hmm in the task Correct order it didn't accept my answer, but it had printed reverse order my code: public static void reverseArray(int[] array) { for (int i = array.length; i > 0; i--){ System.out.print(array[i - 1] + ", "); } System.out.println(); }
Feralny Level 21, Łódź, Poland
27 November 2022
Your code does not reverse the order of numbers, but simply prints the same array with reverse order, and that's the difference
Evan Level 33, United States of America, USA
1 January 2023
To add to what Feralny said, the reverseArray() method must only reverse the array. The main method calls the printArray() method both before and after the reverseArray() method is called.
Abe Level 13, los angeles, United States
21 January 2023
what is the difference of those two?
20 March 2023
The reverseArray() method changes the order of the array elements whether or not they're printed. And, I think, if you display the array in this solution with toString(), you'll see that the elements are not really reordered. It's like you're literally printing a document starting with the last page, then the second to the last, and so on until the first page (descending), but if you view the soft copy, the pages of the document are still in the same order (ascending).
Omar Garnica Level 14, Mexico
25 January 2024
You can always confirm the order of the elements in any array using:

System.out.println( Arrays.toString( array ) );
Try printing your array before and after calling the reverseArray method