Can anyone explain the logic behind it

Question

what would be the output of

1).

     int j=0;
     for (int i=0; i<100; i++) j=j++;
     System.out.println(j);

I thought j=j++; will be equal to

 int j2 = j;
    j = j+1;

so I was expecting the out put would be 99. but when I compiled on eclipse output was 0.

2). and I could not understand what is the logic behind

  ((int)(char)(byte) -1)

When ran it on eclipse I got output as 65535.

1. One question per question please. 2. Try to give your question a name that describes the problem. — Keppil, Sep 18 '14 at 08:44
Duplicates: 1. http://stackoverflow.com/q/2750216/1343161 2. http://stackoverflow.com/q/7500523/1343161 — Keppil, Sep 18 '14 at 08:46
I do not want to be rude, but I think most of these things are nicely explained in the Internet, and in most programming books. If you want to learn a programming language, it is definitely helpful to read a related book. StackOverflow's is a place to discuss more serious (or should I say, advanced) matters. Learn the basics and try again. You may find programmers.stackexchange.com also useful. You should also check the terms of use in the help center for both programmers and SO — Ivaylo Slavov, Sep 18 '14 at 08:56

score 5 · Answer 1 · answered Sep 18 '14 at 08:44

j=j++;

is functionally equal to

int xxx = j;
j++;
j = xxx;

So the value of j stays the same. (Because the right side is evaluated first, including the increment, then the result is assigned to j)

As for ((int)(char)(byte) -1), a char is 16bit in size and unsigned, so the bit pattern of -1 results in 65535.

score 1 · Answer 2 · answered Sep 18 '14 at 08:44

This is because the ++ works as follows:
a = 0;
a = a++; // a will get the assignment of the current value of a before the increment occurs so a = 0 in here

However, in the next case here:
a = 0;
a = ++a; // a will get the assignment of the incremented value of a so a = 1 in here

Can anyone explain the logic behind it

2 Answers2