preincrement / postincrement in java

Question

Can someome help me to understand why:

int i=1;
int j=1;
int k=1;
int l=1;

System.out.println(i++ + i++);  
System.out.println(++j + ++j);  
System.out.pri         


        

Answer 1

As the name indicates, a post increment increments the value of the variable AFTER the variable has been processed (read) while the pre incrment increments the value BEFORE.

For i, that means that first i is incremented by 1, but read as 1, then incremented by 1 again (already being 2 now from the first increment), thus incremented to 3, but read as 2. This results in 1+2 = 3 and the value of i will be 3 as well...

Answer 2

i++ + i++

means use i, then increment, so i is pushed to some kind of stack,

then increased by 1,

then the operator (+) is pushed to the stack,

then i (now 2) is pushed to the stack.

Since the expresseion is now over, the values and operator are popped: the 2nd i is 2, the first i is 1, 2+1=3 (i is now 3, since it was incremented after being pushed).

The thing you are probably missing is that i isn't increased after the evaluation of the whole expression, in the case of a postincrement, and vice versa for preincrement.

Answer 3

Variable++ means: Increment variable AFTER evaluating the expression.

++Variable means: Increment variable BEFORE evaluating the expression.

That means, to translate your example to numbers:

System.out.println(i++ + i++);  //1 + 2
System.out.println(++j + ++j);  //2 + 3
System.out.println(k++ + ++k);  //1 + 3
System.out.println(++l + l++);  //2 + 2

Does this clear things up, or do you need further explanations?

To be noted: The value of all those variables after the 'println' equal '3'.

Since the OP asked, here's a little 'use-case', on where this behaviour is actually useful.

int i = 0;
while(++i < 5) {           //Checks 1 < 5, 2 < 5, 3 < 5, 4 < 5, 5 < 5 -> break. Four runs
    System.out.println(i); //Outputs 1, 2, 3, 4 (not 5) 
}

Compared to:

int i = 0;
while(i++ < 5) {           //Checks 0 < 5, 1 < 5, 2 < 5, 3 < 5, 4 < 5, 5 < 5 -> break. Five runs
    System.out.println(i); //Outputs 1, 2, 3, 4, 5
}

Answer 4

Things to know:
1. Java evaluates expressions from left to right
2. ++i-pre-increment i.e increment before assignment
3. i++ - post increment i.e. increment after assignment

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

op1=i++
op2=1++
sum=op1+op2
i++ - post increment the value of i

1. Assign i to op1 and then increment the value of i.op1=1,i=2
2. Assign i to op2 and then increment the value of i.op2=2,i=3

3. Sum=3

   System.out.println(++j + ++j);

op1=++j
op2=++j
sum=op1+op2
++i - Pre increment the value of i

1. now at first j will be incremented to 2 and then assigned to op1.
2. Then, j will again be incremented to 3 and assigned to op2

3. Then, op1 and op2 will be added to print the sum as 5 and the value of j will be

   System.out.println(k++ + ++k);

op1=k++
op2=++k
sum=op1+op2

1. Assign the value of k to op1 and then increment k. op1=1,k=2

2. Increment the value of k and then assign to op2. op2=3,k=3

3. Sum = 4

   System.out.println(++l + l++);

Apply the above logic here also.

Answer 5

System.out.println(i++)

-> produces 1 since it increments after printing, but when call it twice u will have 1 + 2 so u can translate it to

System.out.println(int i=i+1, plus int i = i + 1 -> this gives 2)

fist phrase gives 2 and second gives 3 but u print it before they i++ increments , so u will have 1 + 2 at the end

System.out.println(++j);

-> produces 2 since it increments before printing

So when u will have ++j = 2 and then ++j = 3 so ++j and ++j is now 5