3

I'm struggling to figure out why the below Java code has the following output.

Specifically, why does the value of s get set to 125 each time and the value of ns get set to 0 for ref3?

Code

public class App {

  int ns;
  static int s;

  App (int ns) {
    if (s < ns) {
      s = ns;
      this.ns = ns;
    }
  }

  void doPrint () {
    System.out.println("ns = " + ns + ", s = " + s);
  }

  public static void main(String[] args) {
    App ref1 = new App(50);
    App ref2 = new App(125);
    App ref3 = new App(100);

    ref1.doPrint();
    ref2.doPrint();
    ref3.doPrint();
  }
}

Output 

ns = 50, s = 125
ns = 125, s = 125
ns = 0, s = 125
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1
  • 100 is less than 125, so your if is false and the assignments are not executed (and ns stays at the initial default). Commented Jul 27, 2016 at 1:18

5 Answers 5

Reset to default
4

There are different aspects that come into play.

Default values

Variables in Java have default values, even if you don't specify them. For int this is 0, so the initial values of your App class are:

int ns = 0;
static int s = 0;

So actually ns is not "set to 0 for ref3", but instead just keeps its initial value, whereas ref1 and ref2 explicitly alter the value.

Variable scopes

Let's look at the constructor of App:

App (int ns) {
  if (s < ns) {
    s = ns;
    this.ns = ns;
  }
}

There are two different variables with the name ns:

  • this.ns, which references the field on instance level (which at this point still has its initial value 0)
  • ns (without this), which is only visible inside the constructor and has the value provided via the constructor argument (50, 125 or 100).

Static variables

s is a static variable and associated with the App class itself, not with a specific instance. So whenever one instance changes its value (e.g. because 50 < 125 in the constructor), all other instances (in fact: the entire program) will see the updated value.

The logic inside the constructor basically sets s to the maximum of all provided ns values so far, which is in this case 125. Try to move your doPrint() methods a few lines above and see how those values change.

Appendix: "Debugging"

For a complete overview of how the values evolve you can also add some "debug" statements:

System.out.println("App.s=" + App.s);
App ref1 = new App(50);
System.out.println("ref1.ns=" + ref1.ns + ", ref1.s=" + ref1.s + ", App.s=" + App.s);
App ref2 = new App(125);
System.out.println("ref2.ns=" + ref2.ns + ", ref2.s=" + ref2.s + ", App.s=" + App.s);
App ref3 = new App(100);
System.out.println("ref3.ns=" + ref3.ns + ", ref3.s=" + ref3.s + ", App.s=" + App.s);

This will output:

App.s=0
ref1.ns=50, ref1.s=50, App.s=50
ref2.ns=125, ref2.s=125, App.s=125
ref3.ns=0, ref3.s=125, App.s=125

You can see that ref.s is always equals App.s and App.s exists even before you create your first instance.

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Comments

2

ns is set when s < ns,

for ref1 you pass ns = 50, but s is uninitialized and java assumes s as zero, so it enters if ( s < ns ) and sets ref1's class variable ns and also static nsto 50

for ref2 you pass ns = 125 but s is 50, if( s < ns ) holds true, so enters if and updates static variable s to 125 and ref2's class variable ns to 125

for ref3 you pass ns = 100, but s is already 125 and hence doesn't enter if condition, there ns stays uninitialized whose value is zero when printed.

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0

As you know, when a variable is declared with the keyword “static”, its called a “class variable”. All instances share the same copy of the variable.

So when App ref2 = new App(125) has run, s is 125. And in App ref3 = new App(100), ns is 100, leee than s (125), so your if assignments are not executed, which mean ns stays at the initial default 0.

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0

Static in Java roughly means that a variable is shared between all class instances (and class itself). It is initialized before everything else, and when you update it in one instance it's reflected in all other instances.

Your program's execution is as follows:

  • Initialize s to 0 (default int)
  • 0 < 50 therefore set both to 50
  • 50 < 125 therefore set both to 125
  • 125 > 100 therefore don't execute if branch. ns is uninitialized at the end of constructor, so it gets initialized to 0 by default

Also, you are printing values after that steps, so our unique s is bound to 125

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0

The value of s gets to 125 each time because it is a static variable and static variables are common to the class so their values do not change for every object that you create.As you call the doPrint method after all the object creations so the recent value of s is 125 and so is your output for every call to the doPrint method.

The value of ns is set to 0 for ref3 beacause in your constructor you are comparing s and ns .When you create reference variable ref3 this happens in the constructor:

    `App (int ns) 
     {
      if (s < ns)   /* s is 125 and ns is 100 so the if statement evaluates 
                     to  false and the assignments don't happen*/
     {
     s = ns;        // this isn't performed
     this.ns = ns;  // even this isn't performed
     }
    }`

So when you call ref3.doPrint() value of ns is 0 and s is 125 and so is your output .

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