Java Programming - Language Fundamentals - Discussion
Discussion Forum : Language Fundamentals - Finding the output (Q.No. 1)
1.
What will be the output of the program?
public class CommandArgsThree
{
public static void main(String [] args)
{
String [][] argCopy = new String[2][2];
int x;
argCopy[0] = args;
x = argCopy[0].length;
for (int y = 0; y < x; y++)
{
System.out.print(" " + argCopy[0][y]);
}
}
}
and the command-line invocation is
> java CommandArgsThree 1 2 3
Answer: Option
Explanation:
In argCopy[0] = args;, the reference variable argCopy[0], which was referring to an array with two elements, is reassigned to an array (args) with three elements.
Discussion:
36 comments Page 4 of 4.
Anita said:
1 decade ago
Here argcopy declared within 2 dimensional array.
Arun kumar said:
1 decade ago
Here argCopy[0](reference of one dimensional array) is referring args. This means, argCopy[0] is pointing the memory location of args. Thats why it prints the values which are stored in args.
Santoshkumar said:
1 decade ago
Args is an array declared in main method which will takes the command line arguments and store in an array. That will be stored in argval[0]. So here array will be stored in another element of array so now that element will acts as two dimensional array. That thing we are printing.
Santosh kumar said:
1 decade ago
Argcopy declared with in two dimensional array but args initialised to argcopy[0]. Is it correct. Please explain.
Lachya S said:
1 decade ago
For me code compile successfully but when I was giving 1 2 3 as an command argument it is throwing error like 1 is not recognised as an internal or externalcommand. what should i do? Please give me solution for that issue.
Abhijit said:
1 decade ago
Can anyone explain how is it come?
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