Java Programming - Operators and Assignments - Discussion
Discussion Forum : Operators and Assignments - Finding the output (Q.No. 7)
7.
What will be the output of the program?
class Test
{
public static void main(String [] args)
{
int x= 0;
int y= 0;
for (int z = 0; z < 5; z++)
{
if (( ++x > 2 ) && (++y > 2))
{
x++;
}
}
System.out.println(x + " " + y);
}
}
Answer: Option
Explanation:
In the first two iterations x is incremented once and y is not because of the short circuit && operator. In the third and forth iterations x and y are each incremented, and in the fifth iteration x is doubly incremented and y is incremented.
Discussion:
31 comments Page 3 of 4.
Nataliia said:
10 years ago
class Main {
public static void main(String[] args) {
int x = 0;
int y = 0;
for (int z = 0; z < 5; z++) {
System.out.println("Before if: " + x + " " + y);
if ((++x > 2) && (++y > 2)) {
System.out.println("Before inc: " + x + " " + y);
x++;
System.out.println("After inc: " + x + " " + y);
}
System.out.println("After if: " + x + " " + y);
}
System.out.println("Result : " + x + " " + y);
}
}
Console:
Before if: 0 0
After if: 1 0
Before if: 1 0
After if: 2 0
Before if: 2 0
After if: 3 1
Before if: 3 1
After if: 4 2
Before if: 4 2
Before inc: 5 3
After inc: 6 3
After if: 6 3
Result : 6 3
public static void main(String[] args) {
int x = 0;
int y = 0;
for (int z = 0; z < 5; z++) {
System.out.println("Before if: " + x + " " + y);
if ((++x > 2) && (++y > 2)) {
System.out.println("Before inc: " + x + " " + y);
x++;
System.out.println("After inc: " + x + " " + y);
}
System.out.println("After if: " + x + " " + y);
}
System.out.println("Result : " + x + " " + y);
}
}
Console:
Before if: 0 0
After if: 1 0
Before if: 1 0
After if: 2 0
Before if: 2 0
After if: 3 1
Before if: 3 1
After if: 4 2
Before if: 4 2
Before inc: 5 3
After inc: 6 3
After if: 6 3
Result : 6 3
Peacz said:
1 decade ago
@Uma you can see @Narayana Murthy explain.
He explain so clear.
He explain so clear.
Uma said:
1 decade ago
Explain me clearly I did not get?
Arnold villasanta said:
1 decade ago
z ++x ++y x++ x y
0 executed not executed not executed 1 0
1 executed not executed not executed 2 0
2 executed executed not executed 3 1
3 executed executed not executed 4 2
4 executed executed executed 6 3
5 (loop stop)
ROhIt said:
1 decade ago
Hi,
What will happen ++(increment operator) preceding & proceeding the variable?
What will happen if in above sum it is :
if (( x++ > 2 ) && (y++ > 2))
Instead of,
if (( ++x > 2 ) && (++y > 2))
And I think the solution described by @Saranya is wrong as it will compare the variable & after that it will increment the variable(i.e after the iteration is completed).
What will happen ++(increment operator) preceding & proceeding the variable?
What will happen if in above sum it is :
if (( x++ > 2 ) && (y++ > 2))
Instead of,
if (( ++x > 2 ) && (++y > 2))
And I think the solution described by @Saranya is wrong as it will compare the variable & after that it will increment the variable(i.e after the iteration is completed).
Thanvir.cse said:
1 decade ago
if (( ++x > 2 ) && (++y > 2)) here x's and y's incrementation does not depend on "if" condition.
When(until for loop condition is true) execute that block i mean "if (( ++x > 2 ) && (++y > 2))" then always x will be incremented.
While first condition is true then check second one(as rules of Short circuit AND (&&) ).
That means y will be incremented. And when for(int z=4;4<5;z++)->if((5>2)&&(3>2)){ x++ /* x=5+1*/ }.
When(until for loop condition is true) execute that block i mean "if (( ++x > 2 ) && (++y > 2))" then always x will be incremented.
While first condition is true then check second one(as rules of Short circuit AND (&&) ).
That means y will be incremented. And when for(int z=4;4<5;z++)->if((5>2)&&(3>2)){ x++ /* x=5+1*/ }.
Vinoth said:
1 decade ago
I'm still confusing because "if" condition is failed then the x would not be incremented, It would come out from the loop. Explain me please?
Saranya said:
1 decade ago
for(int z=0;0<5;z++)->if((1>2)&&(0>2))
for(int z=1;1<5;z++)->if((2>2)&&(0>2))
for(int z=2;2<5;z++)->if((3>2)&&(1>2))
for(int z=3;3<5;z++)->if((4>2)&&(2>2))
for(int z=4;4<5;z++)->if((5>2)&&(3>2))
Now x=5
So,x++=6
Now x=6 y=3.
for(int z=1;1<5;z++)->if((2>2)&&(0>2))
for(int z=2;2<5;z++)->if((3>2)&&(1>2))
for(int z=3;3<5;z++)->if((4>2)&&(2>2))
for(int z=4;4<5;z++)->if((5>2)&&(3>2))
Now x=5
So,x++=6
Now x=6 y=3.
Narayana murthy said:
1 decade ago
@Sudhanshu
The problem is with the two conditions in "if".
If one condition fails then "if" never go for the other condition.
In the example:
In the first iteration, it goes in to the "if" and first increments "x" and checks the condition.. it fails. then it wont go for the next condition to check(we used "&&" so both the conditions should be true to enter into the "if" block.
next iteration will also fails (Coz 2>2 fails)
In third iteration "x" will be 3. but "Y" incremented by 1 so "y=1"(again fails 1>2)
In fourth iteration "x" will be 4 but still the second condition is not satisfied "y=2" only (2>2 fails).
In fifth iteration "x" will be 5 and "y" will be 3, condition true so enters into "if" and increments x by 1. so x=6.(6>2 and 3>2)
finally x=6 and y=3..
The problem is with the two conditions in "if".
If one condition fails then "if" never go for the other condition.
In the example:
In the first iteration, it goes in to the "if" and first increments "x" and checks the condition.. it fails. then it wont go for the next condition to check(we used "&&" so both the conditions should be true to enter into the "if" block.
next iteration will also fails (Coz 2>2 fails)
In third iteration "x" will be 3. but "Y" incremented by 1 so "y=1"(again fails 1>2)
In fourth iteration "x" will be 4 but still the second condition is not satisfied "y=2" only (2>2 fails).
In fifth iteration "x" will be 5 and "y" will be 3, condition true so enters into "if" and increments x by 1. so x=6.(6>2 and 3>2)
finally x=6 and y=3..
Sudhanshu said:
1 decade ago
Please explain this example practically.
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