#include<iostream.h>
void main()
{
int i = 10;
clrscr();
cout << i << i++ << ++i;
getch();
}
The output is
11 12 12
how these output is possible? plz give details explain
6  9235 
The provided code will not produce the output you have given. The correct output should be:
(1) 10,original value, followed by,
(2) 10, the second value. Here, the ++ is executed after the insertion operator.
Now at the beginning of the third increment you will have 11.
(3) At this time, the ++ is executed before insertion operator works. so you will get 12.
In short the output should be 10 10 12.
BTW, what is the point of having getch here???
Banfa 9,065
Expert Mod 8TB
Well I actually get 12 11 12
vmohanaraj I am afraid you are wrong in your explaination
The line of code
cout << i << i++ << ++i;
Invokes undefined behaviour, that is calling this line of code allows the compiler to do anything it wants to. This is because you have altered a single variable (i) more than once between sequence points.
There is no point trying to explain the output, it is very possible that it will be different on every platform that you try it on.
Oops. It is really confusing. I really got 10 10 12, the way I thought it should be. So i went on explaining in that way. I am sorry for that.
However, what is the reason for the undefined behaviour? The statements are executed from left to right. I think that is true for << operator as well. That is why we get the output in that order. Therefore, is not it reasonable to expect that all of us should have got atleast 10 as the first number?
Banfa " Well I actually get 12 11 12" can you explain how this output is evaluated?
i got 12 11 11. i used a printf statement.
printf(" %d %d %d" ,i,i++,++i);
can you explain how the printf will be evaluated?
I Thnk u guys are wrong
In c++, it works from right to left associativity.....thts why we get the answer as 121111
But in C, not in the way
The standard does not mandate any specific behaviour when you modify a
value twice between two sequence-points. Since no specific behaviour is
mandated, the behaviour you see is okay. Note however that that has
nothing to do with the increment-operator in particular.
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