A Doubt

[Niklas L]

No need to be confused ++i => i = 3 ++i => i = 4 + => 4+4 = 8

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[Chris Losinger]

where does the 2nd '4' come from, in your bottom line ?

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[Niklas L]

There is only one instance of i, so i will be incremented twice before the addition occurs.

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[Chris Losinger]

:-O

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[Aescleal]

Just playing devil's advocate... But you're not adding i twice, you're adding the results of the expression ++i which is either going to be 3 or 4 depending on how many times you've done the ++ operation. Cheers, Ash

[Niklas L]

...but if you add 3 and 4 you will not get 8, and the OP would not have been confused in the first place. It would be a shame if once in a while you didn't get an unexpected result. (The testing department might disagree on that though)

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[Lost User]

This is a common issue and the rules are: don't do this as the results are not guaranteed. Whatever answer you get you may get a different answer with another manufacturer's compiler. If you really need to evaluate such an expression you need to break it into steps like:

int i = 2;
int j = ++i;
j += ++i;

It's time for a new signature.

[Stephen Hewitt]

It's pretty subtle. Here's a breakdown of what's hapenning:

1. ++i increments i (so i=3) and returns a reference to i.
2. The second ++i increments i (so i=4) and returns a reference to i.
3. The + is called with the two references. Both are dereferenced and added (4+4).

Steve

[Tim Craig]

To reinforce what Richard said, while this is cute, it's always counterproductive. It's so subtle that even if you KNOW what you're doing, you're going to screw it up and in a production environment, you're going to confuse someone who's not was good with obfurscated code as you think you are. Just write it out in the clearest possible terms and let the compiler worry about optimizing it these days. The compiler can do optimizations that would curl your hair if you looked at what it was doing. :)

Once you agree to clans, tribes, governments...you've opted for socialism. The rest is just details.

[JobinG]

int i = 2; int j = ++i + i++ + ++i + i++ + ++i; cout<<"Value of j: "<<j; The value getting for 'j' is 19. As you have said the three ++i statements will execute and the value of i become 5. How will execute the i++ statements? Could you please explain.

[Stephen Hewitt]

Firstly there are two cases here, the pre-increment and (++i) the post-increment (i++). The pre-increment case increments the variable and returns a reference to it. The post-increment case copies the variable, increments the original and returns the copy.

Steve