i++ and ++i

How will you best differentiate the two?

i++; //and
++i;

++i increments i and then uses the new value, while i++ uses the old value and then increments i.

Quote:

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Originally Posted by alyeska

How will you best differentiate the two?

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By looking at them.

Perhaps you would like to reword your question?

eg.

Code:

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int i = 10;
cout << i-- << endl; //prints 10 (returned THEN decremented)
cout << --i << endl; //prints 8 (decremented THEN returned)

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Code:

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int i = 0;

//first example, b will be 1
int b = ++i;

//second example, b will be 0
int b = i++;

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Note also that in C++, when you are using --x or ++x for your own class (e.g. an iterator), then the calculation is simpler than for x++ or x--, because the latter requires a temporary copy of the value to be returned, and this adds extra work for the compiler. This has been discussed quite a lot - many C++ programmers therefor prefer to use --x or ++x whenever possible, even on standard base-types that aren't subject to custom operator functions [and thus have no difference in performance].

--
Mats

Quote:

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Originally Posted by cyberfish

eg.

Code:

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int i = 10;
cout << i-- << endl; //prints 10 (returned THEN decremented)
cout << --i << endl; //prints 8 (decremented THEN returned)

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On the last part, it will print 9. Just because there are two -'s doesn't mean it will decrement by one. Which one to use? Depends on what you're programming.

Actually, it will print 8, because just before that point i=9.

Uh? For starters you are worng. It prints 8. Next, did you really have a question or are you wasting our time. You seem to know the answer.

Quote:

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Originally Posted by laserlight

Actually, it will print 8, because just before that point i=9.

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Oops, I saw that statement by itself and the declaration of the variable. So I guess it would print 8. Good catch :P