Java Programming - Language Fundamentals - Discussion

@All.

Here I got the Output :
C:\Users\user\OneDrive\Desktop>javac X.java

C:\Users\user\OneDrive\Desktop>java X Java X a b
a
Note : Entering Four Elements then args.length =4 and the loop iterates from o to3. names[0]=args[0];// Java
names[1]=args[1]; // X
names[2]=args[2]; // a
names[3]=args[3]; // b
we are printing names[2]; // a is the output.

The output will be b.

@Jack: public static void main(String [] args).

Here, args is a reference variable of type String array type. It's like void display(int a): here a is "a" is a variable of type "int".

args.length() is a pre defined method which tells about the length of the array which is defined.

Ex: String names[] = new String[5]; here, length of array is 5.

What is args.length? What is there in args?

Hello every using notepad the above code works.

But "using Eclipse IDE" as per below code output is as follows.

I don't understand how to assign a, b parameters to names[0] and names[1]. Please help me in solving this, using Eclipse IDE.

public class X
{
public static void main(String [] args)
{
String names [] = new String[5];
for (int x=0; x < args.length; x++)
names[x] = args[x];
System.out.println(names[0]);
System.out.println(names[1]);
System.out.println(names[2]);
}
}

Output:

null.
null.
null.

@Ritesh.

In this question we have already declared a string of size 5. Hence the values are a, b, null, null, null.

And in that 2nd question of that link string is not declared and if we assign value that does not exist it gives Array out of bound Exception.

Why an ArrayOutOfBound exception is not thrown in this case?

As args[2], args[3], args[4] does not exist and we are assigning these value to name[x].

Please see the following link. In ques.2, this concept is used.

http://www.indiabix.com/java-programming/language-fundamentals/002001

Please someone solve my confusion.

Taking the name of the program in the arguments is seen in C. Java on the other hand doesn't do that. Check this with any parameters you want.

public class Mpla {

public static void main(String [] args)
{

for (int y = 0; y < args.length; y++)
{
System.out.print(" " + args[y]);
}
}
}

It will not print Mpla.

Even X is class but it has been declared in command line so how could it be not recognised.

X is class. hence its java X, a and b are parameters who assigned as 0th n 1st elemnts n others as null..