++i vs i++; what's the diff??

[sballew]

I am confused on the logic of expressions i++ and ++i and --i

[code]
int i, j, k;

i=3, j=4, k=5;
printf("%d ", i++ - j++ + --k); //this one is the confusing part//
printf("%d %d %d, i, j, k); // the answr depndt on 1st part//

[\code]

the above yields following output:

3 4 5 4


Why??

first integer printed = 3; why??

i= 3 j=4 k=5
i++ I am thinking i = 3+1 =4
j++ I am thinking j = 4 + 1 = 5
--k I am thinking k = -1 + 5 = 4

So with these assumptions:
i++ - j++ + --k =
4 - 5 + 4 = -5
but when I input code and compile; I get 3 HOW ??

[Stoned_Coder]

the difference between ++i and i++ depends on whether or not you are passing on to another expression.

for instance....

int i=0;
i++;
is the same as
int i=0;
++i;

but when they combine with other expressions this happens...

++i is called prefix increment. This means 1 is added to i and then that value is passed on.
i++ is called postfix increment.This means the value of i is passed on and then i is incremented.

-- works the same.

[sballew]

How about evaluating the first of the samples I gave
so I understand how i++ postincrements and ++i
preincrements

also --i ; is this subtracting 1 before passing i???

[Stoned_Coder]

ok here:-
printf("%d ", i++ - j++ + --k);

lets say i is 5 j is 3 and k is 2

i++ is pass 5 then increment i to 6
- j++ subtract 3 from 5 then increment j to 4
+--k subtract 1 from k then add so this evalutes to +(2-1)

so the whole expression 5-3+1=3

[sballew]

I get it now.

Wish the textbook explained as well as you did, Stone.