CODEWITHSUNDEEP
javafor-loop
user2770838
user2770838

Reputation:

Unsure about for-loop iteration?

So I'm new to programming in Java and I'm just having a hard time understanding why this

for (int i = 0, j=0; i <10; i++) {
    System.out.println(j += j++);
}

prints out 0 ten times?

Upvotes: 1

Views: 115

Answers (4)

user2577576
user2577576

Reputation:

Are you unsure about the for-loop? int i = 0 declares an int i, and sets it to 0. j = 0 also declares another int j, and sets it to 0. i < 10 states that while i is less than 10, the loop will keep on going. Finally, i++ states that every time the loop is done, i = i + 1, so essentially one will be added to i.

Upvotes: 0

arshajii
arshajii

Reputation: 129587

j += j++

can be thought of as

j = j + j++

Now, we start with j = 0, so j++ increments j and returns its old value of 0 (!), hence we essentially are left with

   j = 0 + 0
//     ^   ^
//     j   j++

ten times. The incrementation of j is overriden by the fact that we reassign j to the outcome of the right hand side (0) just after.

Sometimes I find it helpful to look at the bytecode. j += j++ is really:

ILOAD 1    // load j, which is 0
ILOAD 1    // load j, which is 0
IINC 1 1   // j++ 
IADD       // add top two stack elements
ISTORE 1   // store result back in j

Since IINC does not alter the stack in any way, IADD adds the value of j to itself: 0 + 0. This result is stored back into j by ISTORE after j has been incremented by IINC.

Upvotes: 5

Pshemo
Pshemo

Reputation: 124275

In j += j++ you are actually doing 

j = j + j++;

so for j=0 you will get 

j = 0 + j++

and since j++ will increment j after returning its value you will get

j = 0 + 0;

for now after j++ j will be equal to 1 but, after calculating 0+0 it will return to 0 and that value will be printed.

Upvotes: 1

newuser
newuser

Reputation: 8466

System.out.println(++j);

instead of

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

Upvotes: -1

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