Local (block) variables cannot have the same name with variables declared in outer blocks. Only one declaration of x is permitted. Even if the x variable of the main block is renamed to y, still there is a problem. If the variable x of the first (outer) for block will be renamed to z, only then the program compiles What will happen when you compile this program and then execute it? public class NestedBlocks { public static void main (String[] args) { int x; // local in main }} x = 1 ; System.out.println("Main x " + x ); for (int i = 0; i < 2; i++) { int x = 10; // local in for System.out.println("1st For x " + x) ; x++; for (int j = 0; j < 2 ; j++) { } int x = 0; // local in for x++; System.out.println("2nd For x " + x); System.out.println( "1st For x " + x); System.out.println ( "Main x " + x);
Local (block) variables cannot have the same name with variables declared in outer blocks. Only one declaration of x is permitted. Even if the x variable of the main block is renamed to y, still there is a problem. If the variable x of the first (outer) for block will be renamed to z, only then the program compiles What will happen when you compile this program and then execute it? public class NestedBlocks { public static void main (String[] args) { int x; // local in main }} x = 1 ; System.out.println("Main x " + x ); for (int i = 0; i < 2; i++) { int x = 10; // local in for System.out.println("1st For x " + x) ; x++; for (int j = 0; j < 2 ; j++) { } int x = 0; // local in for x++; System.out.println("2nd For x " + x); System.out.println( "1st For x " + x); System.out.println ( "Main x " + x);
Computer Networking: A Top-Down Approach (7th Edition)
7th Edition
ISBN:9780133594140
Author:James Kurose, Keith Ross
Publisher:James Kurose, Keith Ross
Chapter1: Computer Networks And The Internet
Section: Chapter Questions
Problem R1RQ: What is the difference between a host and an end system? List several different types of end...
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I find these types of questions challenging due to a lack of understanding. Could you please provide a detailed and specific answer to help clarify my confusion?
question:
What will happen when you compile this program and then execute it?
![Local (block) variables cannot
have the same name with
variables declared in outer
blocks.
Only one declaration of x is
permitted. Even if the x
variable of the main block is
renamed to y, still there is a
problem. If the variable x of
the first (outer) for block
will be renamed to z, only
then the program compiles
What will happen when you
compile this program and then
execute it?
public class NestedBlocks {
public static void main (String[] args)
{
int x; // local in main
}}
x = 1 ;
System.out.println("Main x " + x );
for (int i = 0; i < 2; i++)
{
int x = 10; // local in for
System.out.println("1st For x " + x) ;
x++;
for (int j = 0; j < 2 ; j++)
{
}
int x = 0; // local in for
x++;
System.out.println("2nd For x " + x);
System.out.println( "1st For x " + x);
System.out.println ( "Main x " + x);](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F2318e523-864b-4b5c-b102-098ecff48e9d%2Ffbbe5087-9768-4367-bd02-16acfb353318%2Fi12zpb_processed.png&w=3840&q=75)
Transcribed Image Text:Local (block) variables cannot
have the same name with
variables declared in outer
blocks.
Only one declaration of x is
permitted. Even if the x
variable of the main block is
renamed to y, still there is a
problem. If the variable x of
the first (outer) for block
will be renamed to z, only
then the program compiles
What will happen when you
compile this program and then
execute it?
public class NestedBlocks {
public static void main (String[] args)
{
int x; // local in main
}}
x = 1 ;
System.out.println("Main x " + x );
for (int i = 0; i < 2; i++)
{
int x = 10; // local in for
System.out.println("1st For x " + x) ;
x++;
for (int j = 0; j < 2 ; j++)
{
}
int x = 0; // local in for
x++;
System.out.println("2nd For x " + x);
System.out.println( "1st For x " + x);
System.out.println ( "Main x " + x);
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