SCJP Preparation
Ashish Soni
Greenhorn
Posts: 24
posted 8 years ago
All ,
I am going through one of the SCJP Mock test and encountered a question on array and having tough time to understand the result , Please help and explain ..
class SCJPTest
{
public static void main(String[ ] args)
{
int[] x = { 1, 2, 3, 4 };
int[] y = { 2, 3, 1, 0 };
System.out.println( x [ (x = y)[3] ] );
}
}
The ans is 1
I am going through one of the SCJP Mock test and encountered a question on array and having tough time to understand the result , Please help and explain ..
class SCJPTest
{
public static void main(String[ ] args)
{
int[] x = { 1, 2, 3, 4 };
int[] y = { 2, 3, 1, 0 };
System.out.println( x [ (x = y)[3] ] );
}
}
The ans is 1
Roopa Maheshkumar
Greenhorn
Posts: 26
posted 8 years ago
--------------------------------------------------
Please check if this explanation helps
if someone finds this incorrect, please correct me
Consider the left to right evaluation order in x[ (x=y)[3] ];
First the value of outermost x is evaluated which is reference to the first array {1,2,3,4}
Now the assigment is considered x=y which is reference to the second array { 2,3,1,0}
so value of (x=y)[3] is 0 which is the element at index 3 in the second array
Now the value is substituted ...so x[(x=y)[3]] becomes x[0]
Now the value of the x[0] as per the first array is 1
After the execution of this statement, the x and y references both point to the second array only.
To understand better, introduce a print statement
System.out.println( x[ 0] );
after the System.out.println( x[ (x=y)[3] ] );
That will give a different result 2..because now reference x also points to second array after the assignment..
[ May 31, 2008: Message edited by: Meena MeenakshiSubramanian ]
Originally posted by Roopa Maheshkumar:
Hi... i have a little confusion here.
it seems to me that x[0] is 2 since x=y = {2,3,1,0}. Please clarify.
--------------------------------------------------
Please check if this explanation helps
if someone finds this incorrect, please correct me
Consider the left to right evaluation order in x[ (x=y)[3] ];
First the value of outermost x is evaluated which is reference to the first array {1,2,3,4}
Now the assigment is considered x=y which is reference to the second array { 2,3,1,0}
so value of (x=y)[3] is 0 which is the element at index 3 in the second array
Now the value is substituted ...so x[(x=y)[3]] becomes x[0]
Now the value of the x[0] as per the first array is 1
After the execution of this statement, the x and y references both point to the second array only.
To understand better, introduce a print statement
System.out.println( x[ 0] );
after the System.out.println( x[ (x=y)[3] ] );
That will give a different result 2..because now reference x also points to second array after the assignment..
[ May 31, 2008: Message edited by: Meena MeenakshiSubramanian ]
Cheers, Meena
OCPJP 6
niveditha ramana
Greenhorn
Posts: 6
posted 8 years ago
(x=y) is within the scope of x[(x=y)[3] ]
so x[ (x=y)[3] ] will give x[0] first and there scope of x=y is over
and x[0] is now 1
please correct if am wrong
thank you
Originally posted by Roopa Maheshkumar:
class SCJPTest
{
public static void main(String[ ] args)
{
int[] x = { 1, 2, 3, 4 };
int[] y = { 2, 3, 1, 0 };
System.out.println( x [ (x = y)[3] ] );
}}
(x=y) is within the scope of x[(x=y)[3] ]
so x[ (x=y)[3] ] will give x[0] first and there scope of x=y is over
and x[0] is now 1
please correct if am wrong
thank you
with regards<br />niveditha<br /> <br />"The pessimist borrows trouble; the optimist lends encouragement"<br />- William Arthur Ward
