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# Doubt from Whizlab Question

cfish meena
Greenhorn

Joined: Apr 05, 2007
Posts: 19
can anyone please explain me why the output is 3.And x-- should be 0.since it changes to 1 only in next place of x right ,as it is a post increment..Am i right ...Please explain this.

public class sample12 {

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;
}
}
anil kumar
Ranch Hand

Joined: Feb 23, 2007
Posts: 447
Hi

public class sample12 {

static{
int x=5;
}
static int x,y;//Default values are x=0,y=0

public static void main(String[] args) {

x--;
myMethod();//At this statement the value of x is -1 and y =0
System.out.println(x+y + ++x);//here 1+0+2
so output is 3

}
public static void myMethod(){
y=x++ + ++x;// at this statement y=-1+1
so y=0;but x=1;
because here x is incremeted twice
}
}

Thanks

Anil Kumar
V Gala
Ranch Hand

Joined: Aug 06, 2007
Posts: 113
hi
when code enter mymethod the value of x is -1

y=x++ + ++x;
x++->the value of x is read first as -1 and then incremented to 0.
++x->the value of x is first incremented to 1 and then read.
therefore the value of x is 1.
now y= -1+1=0
and x=1
now x+ y+ ++x
first expression are evaluated and then addition take place
++x->the value of x is first incremented to 2 and then read.
now x+y+++x = 1+0+2=3
Amitji Sharma
Greenhorn

Joined: Dec 26, 2006
Posts: 16
public class sample12 {

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

public static void main(String[] args) {

x--; //x=-1 here because before this statement x=0(default value)
myMethod();
System.out.println(x+y + ++x);//here (x+y ++x)(1+0+2)and prints 3

}
public static void myMethod(){
y=x++ + ++x; //here y=-1+1 which is zero; and x becomes 1
}
}

Amit Sharma<br />There is only one proof of ability that's Result.
cfish meena
Greenhorn

Joined: Apr 05, 2007
Posts: 19