Java: Prefix/postfix of increment/decrement operators?

从下面或下面的程序中，为什么上次调用System.out.println(i)打印出值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"
          System.out.println(i);    // "7"
     }
}

---

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 and 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 beautfy of a postfix increment.

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

---

I know this has been answered, but thought another explanation may be helpful.

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

---

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"

---

Think of ++i and i++ as SIMILAR to i = i+1. Although its not SAME. Difference is when actually i will be assigned with the new increment.

in ++i , increment happens immediately.

but if i++ is there increment will happen when program goes to next line.

Look at code here.

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

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

This will result non ending loop . because i will be returned with original value and after the semicolon i will get incremented but returned value has not been. Therefore i will never actually returned as an incremented value.

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