how the program works?

[teo kokos]

Hi to all!! I am beginner in Java language and i want your help.
Can anyone explain me exacthly how the program runs?
Which value have the variables every time?

public class TestApp1
{ 
	static int A = 10, X = 2, Y = 3;
	static void P1 (int A, int X, int Y)
	{
		int TEMP;
		TEMP = A * ++X;
		Y = TEMP * X--;
	}
	static void P2 (int X)
	{
		Y = A * X++;
		Y *= ++X;
	}
	public static void main (String args[])
	{
		P1 (A, Y++, --X);
		System.out.println (X+" "+Y);
		P2 (++Y);
		System.out.println (X+" "+Y);
	}
}

[Wouter Oet]

Sounds like a homework assignment to me. How do you think that it works?

[teo kokos]

I cannot understand the difference between the local variables and which variables cosidered local and the global variables. And why the x has the value 1?

[marc weber]

When a parameter is passed to a method, the variable referring to it is local to that method. This might shadow a variable of the same name.

For example, variables A, X, and Y are all declared at the class level. However, there are also variables A, X, and Y that are local to the method P1. These local variables shadow the class variables.

To help understand how this code works, start with the first line in main...

P1(A, Y++, --X);

Without going any further, answer 2 questions:
1) What are the values passed to P1?
2) After that method executes, what are the values of class variables A, X, and Y?

[teo kokos]

Thank you for the help!
1) I believe the values are A=10, Y=4, X=1
2)The class variables are the same bacause are declared static, isnt'it??

[Sunny Bhandari]

Hi teo,

welcome to Ranch

[marc weber]

teo kokos wrote:... 1) I believe the values are A=10, Y=4, X=1...

You need to consider the difference between pre and post incrementing. Basically, if the increment operator is before the variable (for example, --X), then the value is incremented before being used. If the increment operator is after the variable (for example, Y++), then the value is incremented after being used.

So when you call P1(A, Y++, --X), the value for Y is passed before being incremented, and the value for X is passed after being incremented. So the values passed into the method are 10, 3, and 1.

teo kokos wrote:... 2)The class variables are the same bacause are declared static, isnt'it??

No, "static" is not the same as "final." A static variable "stays" with the class, rather than being associated with a particular instance of that class. But unless the variable is also final, its value can change.

[teo kokos]

I cannot understan why the Y has the value 4 when P1 is executed and why not the value (TEMP=10*2=20 Y=20*2=40) 40?

[marc weber]

teo kokos wrote:I cannot understan why the Y has the value 4 when P1 is executed and why not the value (TEMP=10*2=20 Y=20*2=40) 40?

This is the shadowing I talked about above...

When a parameter is passed to a method, the variable referring to it is local to that method. For example, variables A, X, and Y are all declared at the class level. However, there are also variables A, X, and Y that are local to the method P1. These local variables "shadow" the class variables.

In other words, there are 2 separate variables called Y:

There is a static Y associated with the class, TestApp1, and...

There is also a local Y within the method body of P1.

Add the following println statements to your code to see the difference...

static void P1 (int A, int X, int Y) {
	System.out.println("Starting P1...");
	System.out.println("local A is " + A + ", local X is " + X + ", local Y is " + Y);
	System.out.println("static A is " + TestApp1.A + ", static X is " + TestApp1.X + ", static Y is " + TestApp1.Y);

	int TEMP;
	TEMP = A * ++X;
	Y = TEMP * X--;

	System.out.println("Done with P1...");
	System.out.println("local A is " + A + ", local X is " + X + ", local Y is " + Y);
	System.out.println("static A is " + TestApp1.A + ", static X is " + TestApp1.X + ", static Y is " + TestApp1.Y);
}

Outside of P1, 'Y' simply refers to the static variable. So the println in main uses the static Y (not the Y that's local to P1).

Also note the order of the values passed to P1, because they are "switching" X and Y. The method is defined as P1(int A, int X, int Y); and it's called using P1(A, Y++, --X); So inside P1...

10 (the value of static A) is assigned to the local variable A

3 (the value of static Y before incrementing) is assigned to the local variable X (not Y)

1 (the value of static X after decrementing) is assigned to the local variable Y (not X)

[teo kokos]

WoW!!! Thank you very much marc weber!!! I am grateful!!

[teo kokos]

I am checking again this program and there is something where i get stuck...
I made the program like this:

public class TestApp1  
{ static int A = 10, X = 2, Y = 3;  
  
static void P1 (int A, int X, int Y)  
{  
      
System.out.println("Starting P1...");    
System.out.println("local A is " + A + ", local X is " + X + ", local Y is " + Y);    
System.out.println("static A is " + TestApp1.A + ", static X is " + TestApp1.X + ", static Y is " + TestApp1.Y);    
      
int TEMP;  
TEMP = A * ++X;  
System.out.println("New local X is " +X);   
Y = TEMP * X--;  
System.out.println("New1 local X is " +X);   
System.out.println("TEMP is " + TEMP );   
System.out.println("Done with P1...");    
System.out.println("local A is " + A + ", local X is " + X + ", local Y is " + Y);    
System.out.println("static A is " + TestApp1.A + ", static X is " + TestApp1.X + ", static Y is " + TestApp1.Y);    
}  
  
static void P2 (int X)  
{  
      
System.out.println("Starting P2...");    
System.out.println("local A is " + A + ", local X is " + X + ", local Y is " + Y);    
System.out.println("static A is " + TestApp1.A + ", static X is " + TestApp1.X + ", static Y is " + TestApp1.Y);    
System.out.println("New local X is " +X);   
Y = A * X++;  
System.out.println("New1 local X is " +X);   
Y *= ++X;  
System.out.println("New2 local X is " +X);   
      
System.out.println("Done with P2...");    
System.out.println("local A is " + A + ", local X is " + X + ", local Y is " + Y);    
System.out.println("static A is " + TestApp1.A + ", static X is " + TestApp1.X + ", static Y is " + TestApp1.Y);    
  
}  
public static void main (String args[])  
{  
P1 (A, Y++, --X);  
System.out.println (X+" "+Y);  
P2 (++Y);  
System.out.println (X+" "+Y);  
}  
}  

My problem is in P2 method.
a) Why the local X and Y have the value 5?? :O
b) Why the static Y has value 350? With this logic in P1 method the static Y should have value 160.
Please help me because in 2 days i give exams.

[marc weber]

teo kokos wrote:...My problem is in P2 method.
a) Why the local X and Y have the value 5?? :O
b) Why the static Y has value 350? With this logic in P1 method the static Y should have value 160...

I'm glad to see you're working on this! It's a great way to learn.

The method P2 has a local variable X because X is declared as a parameter: static void P2(int X) {...}. But the method P2 does not have local A or Y variables. So in the P2 method body, X refers to a local variable, but A and Y both refer to the static class variables. That means inside P2, A is the same as TestApp1.A, and Y is the same as TestApp1Y.

Also, note that the value passed to P2 is referred to as X (the local variable). However, when P2 is called in main, it's called by passing the incremented value of the static variable Y.

Before that method is called, Y is 4. So calling P2(++Y) increments Y to 5, and passes that value to P2, where it's assigned to the local variable X. So inside P2, X refers to the local variable X, which is 5, and Y refers to the static variable Y, which is also 5.