static block understanding

[Tanu Gulati]

public class test {

static {
int x = 5;
}
static int x, y;

public static void main(String args[]) {
x--;
myMethod();
System.out.println(x + y + ++x);
}

public static void myMethod() {
y = x++ + ++x;

}

}

Please explain how its output is 3?

[Vijitha Kumara]

Do you have a question ?

[Tanu Gulati]

I forgot to write Question. Now i have updated Question Please explain how its output is 3

[Akanksha Mittal]

Hi Tanu

Please Use Code Tags, makes easier to understand and explain.

  public class test { 

static { 
int x = 5; 
} 
static int x, y; 

public static void main(String args[]) { 
x--; 
myMethod(); 
System.out.println(x + y + ++x); 
} 

public static void myMethod() { 
y = x++ + ++x; 

} 

} 

  

Inside the static block you have x declared, so that makes it local to that block and so not used after that.

Now since x and y are class variables, so they are initialized to 0.
while using x and y in the main method only the x and y having initial values 0 are used and hence

after mymethod(): y=0 & x=1
and in println its (1+ 0 + 2)

Remember Postincrement operator increments the value only after the execution of the current statement (in this case - y = x++ + ++x; )
whereas pre increment modifies the variable before its usage

[Sunny X Narula]

One doubt

Remember Postincrement operator increments the value only after the execution of the current statement (in this case - y = x++ + ++x; )

Before the statement, x is -1 and y is 0.
So as per the explanation above the code must translate to:

y=-1 + ((-1)++)=-1 + 0=-1;
x = x+1 = 1

y should not get the value from x++ (thats what I thought) but it does.
In fact it behaves as if the code was

  y = (x++) + (++x);  

That is the only way y can end up with 1 else it should be -1.

May be I am missing some concept here/I might be confused but that is how it works.
I am guessing this has something to do with priority while evaluating an expression. I think ++ has higher priority than +.
Any references will be helpful.
Thanks in advance for any help.

[Akanksha Mittal]

Whenever you say

  
x=0;
y = x++;  

This will make x=1 & y=0

Similarly,

  
x=0;
y = ++x;  

will make x=1 & y=1


In the scenario mentioned above, Pre Increment will be evaluated first, and post increment evaluated last based on precedence so the expression becomes

  x = -1
++x => x=0
y = x++ + 0 => 0+0 =>0 
So, x=1   

Therefore y=0, x=1 in the method "myMethod"

[Sunny X Narula]

Thanks for that Akanksha
Its clear now.