vector expression evaluation

Hello world

Can someone tell me the following expression evaluation.

int i = 0;
System.out.println (i ++ + ++ I + ++ i);


Please help me...

Prem says:
Ok.. Thanks for this knowledge...

If the operands are evaluated, left right then exit this expression must be of 6 but is 5? How?

No, it must be exactly what it is. If such was not the case, it would mean that there is a HUGE mistake in a very basic part of Java a beginner like you could find, but who somehow don't get it fixed, despite Java used in thousands of existing applications.

int i=0;
System.out.println( i++ + ++i + ++i); 

When we reached the first term (i ++), the value of i is 0. The value of this expression is the value of i, before the increment, if we determine that the value of this expression is 0. Then, we increment I 1. Running so our total is 0 and the value of i is 1.

When we reached the second term (++ i), we increment all first I between 1 to 2. Then, we determine the value of this expression, which is the new value of i is 2. We have therefore 0 + 2 = 2 for our running total, and the value of i is now 2.

When we reached the third mandate (++ i once again), we increment all first I have 2-3. Then, we determine the value of this expression, which is the new value of i is 3. So we have 0 + 2 + 3 = 5 for our total and the value of i is 3.

It's as if we have:

int i = 0;
int j = i++; // j  = 0; i = 1;
int k = ++i; // i = 2; k = 2;
int m = ++i; // i = 3; m = 3;
System.out.println(j + k + m); // 0 + 2 + 3

Edited by: jverd May 15, 2012 02:32

Tags: Java

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