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Fine that you have solved it and thank you for the feedback.
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Hi everyone....this Increment(++) and Decrement(--) operators in C topic is really one of the surprise topic for me as I always get some surprise with their answer. The time I thought I got the concept of this topic I find my self wrong with its next question. I want to know what is its actual concept and whats the best approach to solve inc and dec operator questions. Make me understand with an example. Appreciate your suggestion and help. Thanks
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These operators merely add (inc) or subtract (dec) 1 from the item in question. If the operator is used in the postfix position (i++) then the original value is returned. If the operator is used in the prefix position (--i) then the new value is returned. Thus:
int i = 10;
int result;
result = i++; result = ++i; result = i--; result = --i;
result = ++i + i++;
Veni, vidi, abiit domum
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hey i want to rectify Message send by Richard MacCutchan
int i = 10;
int result;
result = i++;
result = ++i;
result = i--;
result = --i;
result = ++i + i++;
As written the last line gives output undefine i think its wrong .it will give correct output as 23.
If u had checked then u already got the result.
here i will explan u how it's work..
1:from left to right ++i changes i values 10 to 11.
2:then it will add 11 + 11 (as i++ is post increment operator it will execute after expression execution) and gives result as 22
3: then i values becomes 22 and i++ will going to execute and i value changes to 22+1=23
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I'm sorry but it is a well known fact that in expressions like this, the compiler is not constrained to follow the rules as you see them. Using multiple increment decrement operators in a single expression is not guaranteed to produce the result you expect and should not be used.
Veni, vidi, abiit domum
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An undefined behavior often means that the compiler may crash, and if not it behaves like it thinks it should, which isn't necessarily the same on different compilers.
Veni, vidi, caecus | Everything summarizes to Assembly code
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Yes. My comments were an over-simplification, but it is still worth avoiding expressions of that sort.
Veni, vidi, abiit domum
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Richard MacCutchan wrote: but it is still worth avoiding expressions of that sort.
Every respectable Dev must avoid it.
By the way, my comment was intended to be an addition to yours.
Veni, vidi, caecus | Everything summarizes to Assembly code
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Marco Bertschi wrote: my comment was intended to be an addition to yours.
Yes, that was how I understood it. 
Veni, vidi, abiit domum
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sukanta kumar mangal wrote: As written the last line gives output undefine i think its wrong .it will give correct output as 23.
If u had checked then u already got the result.
here i will explan u how it's work..
At a minimum, you might want to brush up on sequence points.
"One man's wage rise is another man's price increase." - Harold Wilson
"Fireproof doesn't mean the fire will never come. It means when the fire comes that you will be able to withstand it." - Michael Simmons
"Show me a community that obeys the Ten Commandments and I'll show you a less crowded prison system." - Anonymous
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I'll explain why the behavior is undefined...
This has a lot to do with operator precedence[^], see the ++ as a prefix operator has an equal precedence to the + operator. Its up to the compiler to determine which one to evaluate first, since they are at an equal precedence there is no universal "++(pre) before +".
However, ++ as a postfix operator has a higher precedence than the prefix operator, so for example:
i = 10
result = ++i + i++;
Turns into
result = ++i + 11;
Now, the compiler really doesn't care which one is evaluated first, the + or the ++. Each one is free to implement it as they see fit, so it might be 22 or it might be 23, it depends on which one the programmer handled first.
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no no..in any compiler prefix operator execute first than postfix.
postfix will execute after value returned or assigned to some variable..
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Am I the only one to spot that at least one of the problems in your last line is not related to increment and decrement at all? The operator+ which is invoked here needs to evaluate both of its operands, but it's undefined whether the first or second operand is evaluated first - the compiler may choose to do either!
So, if the compiler evaluates the first operand first, the following happens:
i = 10;
i = i + 1; first_arg = i;
second_arg = i;
i = i + 1; result = first_arg + second_arg;
Now consider the same on the assumption the compiler evaluates the second operand first:
i = 10;
second_arg = i;
i = i + 1; i = i + 1; first_arg = 1;
result = first_arg + second_arg;
In this case, the results are the same, but that is just by coincidence - if you had used subtraction in stead of addition you'd be in for a surprise!
You may want to check http://en.cppreference.com/w/cpp/language/eval_order[^] for further information regarding both function argument evaluation and increment/decrement.
GOTOs are a bit like wire coat hangers: they tend to breed in the darkness, such that where there once were few, eventually there are many, and the program's architecture collapses beneath them. (Fran Poretto)
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That was exactly my point.
Veni, vidi, abiit domum
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You are making unwarranted assumptions! 
It's more complex that that - Richard is right, you shouldn't mix 'em - see my post below for some compiler nasties...
Never underestimate the power of stupid things in large numbers
--- Serious Sam
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I know increment and decrement operator depend on compiler. actually i had tried it in dev c++.
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If you know that something is compiler dependant, then you can't say "it is like this" - because there is a very good chance that the other person is not using the same compiler!
Never underestimate the power of stupid things in large numbers
--- Serious Sam
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int i = 0;
i++;
Means that i is incremented after the statements execution
++i;
Means that i is incremented before the statements execution
The same is applicable on i-- and --i
Veni, vidi, caecus | Everything summarizes to Assembly code
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O God....wow I am not the only one who is confused? This is the problem I specified when I started this thread, I got surprising answers when i solve these inc and dec problems.So who is correct please tell me and whom explaination I go with to pursue my learning on this topic?
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Pre- and post- fix increment and decrement operations are pretty easy in theory, it's only when people get creative that you get problems in practice.
Basically, a prefix (++i or --i) says to increase or decrease the value before you use the variable, so the variable has the new value immediately:
i = 10;
x = ++i + 5; Can be read as:
i = 10;
i = i + 1;
x = i + 5; and similarly for the -- version:
i = 10;
x = --i + 5;
Can be read as:
i = 10;
i = i - 1;
x = i + 5;
The postfix version (i++ or i--) does the same thing, but after the variable has been used:
i = 10;
x = i++ + 5; Can be read as:
i = 10;
x = i + 5;
i = i + 1; And similarly
i = 10;
x = i-- + 5;
Can be read as:
i = 10;
x = i + 5;
i = i - 1;
The trouble comes when you start mixing operations on the same line (as Richard said):
i = 10;
x = ++i + i++; The problem is that the language specification does not define exactly when pre- and post- fix operations should occur! Which means that it's implementation specific exactly what you get as a result: The value of i should always be the same: 12 but the value of x can be different depending on the compiler (and to an extent on the target processor - ARM for example has built in pre- and post- fix increment and decrement to it's "machine code" LOAD operations, so it would be quite likely that an efficient compiler would use them directly)
Should it be executed as:
i = 10;
i = i + 1;
x = i + i;
i = i + 1; Which gives the result 22
or as
i = 10;
i1 = i;
i = i + 1;
x = i1 + i;
i = i + 1; Which gives 21
Or as
i = 10;
i1 = i;
i = i + 1;
x = i + i1;
i = i + 1; which also gives 21 by a different route
And bear in mind that the compiler does not have to evaluate the two operands of "+" in left to right order, so it could even give some very strange and unexpected results! Like 23...
So avoid combining them: use them for "simple expressions" such as incrementing an array index each time round a loop, but don't get fancy, or your code may well fail in interesting ways...
Never underestimate the power of stupid things in large numbers
--- Serious Sam
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Thanks sir for highlighting its deep concepts related to the compiler. I hope now I am able to solve these question without getting shocked by their answers. Appreciate all your responses.
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You're welcome!
Never underestimate the power of stupid things in large numbers
--- Serious Sam
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