K&B Book for Scjp 6 - Chapter 5 question 8 doubt

[Prithvi Sehgal]

Dear All,

I am copying this code from Kathy and Sierra book for SCJP 6. It is from chapter 5. Self test question 8

 
public class Ebb {
	
	static int x = 7;
	
	public static void main(String[] args) {
		String s ="";
		
		for(int y = 0 ; y < 3 ; y++){
			x++;
			switch(x){
			case 8:
				s += "8 ";
			case 9:
				s += "9 ";
			case 10:
				{s += "10 ";break;}
			
			default:
				s+= "d ";
			
			case 13:
				s += "13 ";
			
				
			}
		}
		System.out.println(s);
	}
	
	static{
		x++;
	}

}

I completely understand the code and it's logic but i am unable to understand why the value of x at first iteration used is 9 rather then 8.
To me if you are post incrementing, it means first use the previous value and then increment. Here following sequence i understood.

1- x was initialized with 7
2- As class loaded static initializer block is executed and x is post incremented to 8.
3- After this point inside the loop where x is again incremented, i think x should be incremented to 9 which happens but shouldn't first value
8 be used and then in next iteration value 9 used.

I am sure, i am making some very silly mistake. But confused

Best Regards,

[Srinivas Nagesh]

I havent reached that part in the book but I think this is what happens when you execute the program

1. x is initialized to 7 in line 4
2. x is incremented to 8 in line 32
3. x is incremented to 9 in line 10
4. when the switch stmt is executed for the first time, the value of x is 9. which means case 9 would be resolved.

I am not sure why would you say x would have a value of 8 in the first iteration of the loop?

[Larry Chung]

Prithvi Sehgal wrote:
3- After this point inside the loop where x is again incremented, i think x should be incremented to 9 which happens but shouldn't first value
8 be used and then in next iteration value 9 used.

Perhaps you were thinking about the loop behavior of a for statement that looks like this:

 
       for(int y = 0 ; y < 3 ; y++, x++){     // The x++ is here now
          //x++;                                         // not here
            switch(x){  
            case 8:  
                s += "8 ";  
            case 9:  
                s += "9 ";  
            case 10:  
                {s += "10 ";break;}  
              
            default:  
                s+= "d ";  
              
            case 13:  
                s += "13 ";  
              
                  
            }  
        }     

This code will postpone the increment to after the first iteration is done.

[Prasad Kharkar]

@prithvi
friend there is very very simple solution whenever we want to find out answers like this
why don't you print out the value of x whenever you are confused at what line what is happening
see the following code
that will explain a lot
compile and run it

   public class Ebb {  
       
     static int x = 7;  
       
     public static void main(String[] args) {  
         String s ="";  
           
         for(int y = 0 ; y < 3 ; y++){  
             x++;  
			 System.out.println("value of x in for loop now " + x);
             switch(x){  
             case 8:  
                 s += "8 ";  
			 System.out.println("Value of x in switch block now " + x);
             case 9:  
                 s += "9 ";  
			  System.out.println("Value of x in switch block now " + x);
             case 10:  
                 {s += "10 ";
			  System.out.println("Value of x in switch block now " + x);
			  break;}  
               
             default:  
                 s+= "d ";  
				 System.out.println("Value of x in switch block now " + x); 
             case 13:  
                 s += "13 ";  
			  System.out.println("Value of x in switch block now " + x);
                                  
             }  
         }  
         System.out.println(s);  
     }  
       
     static{  
		 System.out.println("Value of x  in static block before incrememting" + x);
         x++;  
		 System.out.println("Value of x in static block after incrementing" + x);
     }  
   
 }    

and about the post increment
you a little misunderstood
see the following example

  class Increment
{
	static int i = 5;
	public static void main(String [] args)
	{
	System.out.println(i++); // here value of i gets printed first and the value is incremented AFTER statement is executed
	// that is, after semi-colon the value of i is incremented
	System.out.println(i); // here we used the incremented value 
	System.out.println(++i);// and here we are using pre- increment hence the value of i gets incremented BEFORE end of statement
	// and thus we get the value as 7
	}
}  

[Stephen Davies]

Srinivas Nagesh wrote:I havent reached that part in the book but I think this is what happens when you execute the program

1. x is initialized to 7 in line 4
2. x is incremented to 8 in line 32
3. x is incremented to 9 in line 10
4. when the switch stmt is executed for the first time, the value of x is 9. which means case 9 would be resolved.

I am not sure why would you say x would have a value of 8 in the first iteration of the loop?

Ignore the loop as a first reason, it is due to the class static initialization order. When a class first loads (not as an instance object) its static members are initialized first, (in this case x is assigned the value 7) next any static blocks are executed, in this case the static block x++ is run. Thus, by the time main runs, x is initialized to 8, hence after the first loop initialization x is once more incremented by the post increment operator, which ads one to the value of x (which is 8 following class loading) so it becomes 9 and then used in the switch statement, hence the output of 9.

Class load statics:

1.Static members
2.Static members (blocks)

[Prithvi Sehgal]

Guys,

I completely understand the logic and everything in this question, my doubt may be i am unable to clarify. Let say for suppose.

  
  class Increment  
 {  
     static int i = 5;
     static int j = 7 ;
     public static void main(String [] args)  
     {  
     
    	 for(int z = 0 ; z < 1 ; z++){
    		 j++;
    		 System.out.println(j++); // Line 1
    	 }
    		 
     }  
     
     static{
    	 j++;
     }
 }   

At line 1, first the value of j is printed which is 9 and then incremented. My question was shouldn't when this block of code is reached

   
for(int y = 0 ; y < 3 ; y++)
{
  x++;
 switch(x){
  
    //some code here    
}
}

Shouldn't in the above code snippet even first value of x which is 8 should be used and and at next iteration 9 to be used.
Or there is some difference between these two statements

x++;
System.out.println(x++); // Line2

Because in line 2 first previous value of x is used and then incremented.

I know i am stuck at a very silly confusion but still any pointers in this regard will be helpful.

[Stephen Davies]

Prithvi Sehgal wrote:Guys,

I completely understand the logic and everything in this question, my doubt may be i am unable to clarify. Let say for suppose.

  
  class Increment  
 {  
     static int i = 5;
     static int j = 7 ;
     public static void main(String [] args)  
     {  
     
    	 for(int z = 0 ; z < 1 ; z++){
    		 j++;
    		 System.out.println(j++); // Line 1
    	 }
    		 
     }  
     
     static{
    	 j++;
     }
 }   

Shouldn't in the above code snippet even first value of x which is 8 should be used and and at next iteration 9 to be used.
Or there is some difference between these two statements


[code]
x++;
System.out.println(x++); // Line2

Because in line 2 first previous value of x is used and then incremented.

I know i am stuck at a very silly confusion but still any pointers in this regard will be helpful.

I think there are two things here at work. First is static initialization:

In your above code : you set the following static variables:

      static int i = 5;
      static int j = 7 ;

so at class load time, before any instance variables, before main is run etc i = 5, j = 7;

In addition you have a static initializer block:

 static{
    	 j++;
     }

This runs following the static variable initialization but still before main is run or any instance objects are created. So at run time you have:

     i = 5
     j = 8
  

Then in your loop, you first increment j again (j++). So J becomes 9 which is what is passed to the switch statement, when main is finally run

   // Pseudo-code
   switch(9)...

  //Thus when your print statement is called 
  System.out.println(j++);

  //prints 9 and then increments J (now 10 if used again)
   

In your second query you ask about the difference in x++ and System.out.println(x++);

There is no difference in the post increment operator, but when chained with the print statement, the print statement will always print out the value of x before incrementation thus;

  int x = 0;   // x = 0;
  x++;  // x = 1;
  System.out.println(x++); // prints 1, -THEN increments to 2
  System.out.println(x) //prints 2

If you wish to print the incremented value of x either use a second print statement as shown use one print statement with the pre-increment operator e.g.;

  int x = 0;   // x = 0;
 
  System.out.println(++x); increments to 1 and then prints x which is 1

[bhanu chowdary]

Prithvi Sehgal wrote:Guys,


At line 1, first the value of j is printed which is 9 and then incremented. My question was shouldn't when this block of code is reached

   
for(int y = 0 ; y < 3 ; y++)
{
  x++;
 switch(x){
  
    //some code here    
}
}

Shouldn't in the above code snippet even first value of x which is 8 should be used and and at next iteration 9 to be used.
Or there is some difference between these two statements

x++;
System.out.println(x++); // Line2

Because in line 2 first previous value of x is used and then incremented.

I know i am stuck at a very silly confusion but still any pointers in this regard will be helpful.

Post increment does not mean that the value will be incremented after the for loop or switch statement.

  int x = 7;
System.out.println(x++);  

The above code will print 7 because when you are printing the value the end of statement is not reached, once the line of execution passes the println statement the value will be incremented to 8.

 int x = 7;
x++;
System.out.println(x); //In this line the value would be 8.   

When it comes to line three the value is already incremented eventhough it is a post increment operator.

[Prithvi Sehgal]

Thanks bhanu, that made a lot of sense.