Java: Prefix/postfix of increment/decrement operators

From the program below or here, why does the last call to System.out.println(i) print the value 7?

class PrePostDemo {
    public static void main(String[] args) {
        int i = 3;
        i++;

        System.out.println(i);    // "4"

        ++i;
        System.out.println(i);    // "5"
        System.out.println(++i);  // "6"
        System.out.println(i++);  // "6"开发者_运维问答;
        System.out.println(i);    // "7"
    }
}

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i = 5;
System.out.println(++i); //6

This prints out "6" because it takes i, adds one to it, and returns the value: 5+1=6. This is prefixing, adding to the number before using it in the operation.

i = 6;
System.out.println(i++); //6 (i = 7, prints 6)

This prints out "6" because it takes i, stores a copy, adds 1 to the variable, and then returns the copy. So you get the value that i was, but also increment it at the same time. Therefore you print out the old value but it gets incremented. The beauty of a postfix increment.

Then when you print out i, it shows the real value of i because it had been incremented: 7.

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Another way to illustrate it is:

++i will give the result of the new i, i++ will give the result of the original i and store the new i for the next action.

A way to think of it is, doing something else within the expression. When you are printing the current value of i, it will depend upon whether i has been changed within the expression or after the expression.

int i = 1;
result i = ++i * 2 // result = 4, i = 2

i is evaluated (changed) before the result is calculated. Printing i for this expression, shows the changed value of i used for this expression.

result i = i++ * 2 // result = 2, i = 2

i is evaluated after the result in calculated. So printing i from this expression gives the original value of i used in this expression, but i is still changed for any further uses. So printing the value for i immediately after the expression, will show the new incremented value of i. As the value of i has changed, whether it is printed or used.

result i = i++ * 2 // result = 2, i = 2
System.out.println(i); // 2

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If you kept a consistent pattern and included print lines for all the values:

int i = 3;
System.out.println(i);    //  3
System.out.println(i++);  //  3
System.out.println(i);    // "4"
System.out.println(++i);  //  5
System.out.println(i);    // "5"
System.out.println(++i);  // "6"
System.out.println(i++);  // "6"
System.out.println(i);    // "7"

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Think of ++i and i++ as similar to i = i+1. But it is not the same. The difference is when i gets the new increment.

In ++i , the increment happens immediately.

But if i++ is there, the increment will happen when the program goes to the next line.

Look at the code here.

int i = 0;
while(i < 10){
   System.out.println(i);
   i = increment(i);
}

private int increment(i){
   return i++;
}

This will result in a nonending loop. Because i will be returned with the original value and after the semicolon, i will get incremented, but the returned value has not been. Therefore i will never actually returned as an incremented value.

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Why wouldn't the variable have been updated?

• Postfix: passes the current value of i to the function and then increments it.
• Prefix: increments the current value and then passes it to the function.

The lines where you don't do anything with i make no difference.

Notice that this is also true for assignments:

i = 0;
test = ++i;  // 1
test2 = i++; // 1

---

System.out.println(i++);  // "6"

This sends println the value I had prior to this line of code (6), and then increments I (to 7).

---

An example of how the actual operators are implemented:

class Integer {
  private int __i;

  function Integer ++() { // Prefix operator i.e. ++x
    __i += 1; // Increment
    return this; // Return the object with the incremented value
  }

  function Integer ++(Integer x) { // Postfix operator, i.e., x++
    __i+=1; // Increment
    return x; // Return the original object
  }
}

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Maybe you can understand better Prefix/postfix with this example.

public class TestPrefixPostFix
{
    public static void main (String[] args)
    {
        int x = 10;
        System.out.println((x++ % 2 == 0) ? "yes " + x: " no " + x);
        x = 10;
        System.out.println((++x % 2 == 0) ? "yes " + x: " no " + x);
    }
}

---

This may be easy to understand:

package package02;

public class C11PostfixAndPrefix {

    public static void main(String[] args) {
        // In this program, we will use the value of x for understanding prefix 
        // and the value of y for understaning postfix. 
        // Let's see how it works. 
        
        int x = 5; 
        int y = 5; 
        
        Line 13:   System.out.println(++x);  // 6   This is prefixing. 1 is added before x is used. 
        Line 14:   System.out.println(y++);  // 5   This is postfixing. y is used first and 1 is added. 
        
        System.out.println("---------- just for differentiating");
        
        System.out.println(x);  // 6   In prefixing, the value is same as before {See line 13}
        System.out.println(y);  // 6   In postfixing, the value increases by 1  {See line 14} 
        
        // Conclusion: In prefixing (++x), the value of x gets increased first and the used 
        // in an operation. While, in postfixing (y++), the value is used first and changed by
        // adding the number. 
    }
}

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It prints 7 for the last statement, because in the statement above, its value is 6 and it's incremented to 7 when the last statement gets printed.

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This might help... it also took my to understand this puzzle.

public class Main {
  public static void main(String[] args) {

    int x = 5;
    int y = 5;

    System.out.println(++x);
    System.out.println(y++);

    System.out.println("------");

    System.out.println(x);
    System.out.println(y);
  }
}

---

Well, think of it in terms of temporary variables.

i = 3;
i ++; // Is equivalent to:   temp = i++; and so, temp = 3 and then "i" will increment and become i = 4;
System.out.println(i); // Will print 4

Now,

i = 3;
System.out.println(i++);

is equivalent to

temp = i++;  // 'temp' will assume a value of the current "i", after which "i" will increment and become i = 4
System.out.println(temp); // We're printing 'temp' and not "i"