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To really understand why you should realise that statements such as:
i = ++i i = i++
are actually considered very bad form. The preincrement (++i) and postincrement (i++) operators do not need to be part of an assignment statement, and in fact using them as part of one can cause a lot of confusion. For example the statement i=i++ actually has the same result as i=i - which of course is a totally redundant statement. If you change the above two lines to:
or even simply:
i += 2;
you will hopefully see why there is a difference between the two peices of code.
[EDIT: Incidentally, this is the second or third time this problem has come up in recent weeks. Is this just a common beginners mistake or is there actually an institution somewhere giving out this kind of code for assignments?] [ August 04, 2005: Message edited by: Stuart Gray ]
You need to understand the preincrement and postincrement first.
* Preincrement does - first increment the value and then give the value. * Postincrement does - first return(get) the value and then increment. Hence the behavior.
The assignment operation happens like : Say expression is var = (expression); *the expression is evaluated. Say (expression resulted in x) so var = x; *hence var has value x.
If we apply the same to below.
int i=10; int k=0; i = ++i; // ++i results in 11 so i = 11, hence value of i becomes 11 i = i++; // here the i++ result 11 though the value in i is incremented // to 12 then i = 11, hence the value of i becomes 11 System.out.println(i); this will print value as 11
Then below is easy.
and if the same program is written as Int i=10; int k=0; k = ++i; k = i++; System.out.println(i); this will print i as 12
Incidentally, this is the second or third time this problem has come up in recent weeks. Is this just a common beginners mistake or is there actually an institution somewhere giving out this kind of code for assignments?
SCJP study guides always have questions about the behavior of things like
i = i++ + ++i;
as though this were a common sort of programming conundrum. Unfortunately, studying for the SCJP is quite often a fledgling programmer's first exposure to Java.
I am writing a simple program and got confused with a small fragment of code.I am writing that code fragment pls clarify the confusion. int i=10; int k=0; i = ++i; i = i++; System.out.println(i);
this will print value as 11
and if the same program is written as
Int i=10; int k=0; k = ++i; k = i++; System.out.println(i);
this will print i as 12
why is there disprecency in the answers
Hi Sangeeta, There is no discrepancy in the answers. The answers are absolutely correct. Why ? Here's why...
When u say ++i, u can segregate this operation into 3 separate operations : i += 1; result = i; return result;
Thus in the first program, when u say i = ++i, u get the value of i as : i += 1 //i.e..11; result = i //i.e..result = 11 return result; //i.e..return 11;
This value returned (11) is then stored into i. So new value of i becomes 11.
Now when u use the operation i++, u can consider 3 operations again: result = i; i += 1; return result;
So in the next line of the first program, when u say i = i++, u get: result = i //i.e..result = 11 since i got the new value 11 from the previous step. i += 1 //i.e..i = 11+1 = 12. return result //i.e..return 11
Hope u understood that !!
Now coming to the next program lines with variable 'k': Keeping the above 3-line operations in view, we have k = ++i; So, i += 1; //i.e..i=11 result = i; //i.e..result = 11 return result // return 11;
So k = 11.
and the next statement : k = i++ yields
result = i; //i.e..result = 11 since i is 11 from previous step i += 1; //i.e..i = 12; return result; //i.e..return 11
this final value 11 is stored in k. Hence i=11 and k=11 from first step and i=12 and k=11 from second step.
Hope this gets ur fundas absolutely crystal clear !!