Very interesting problem, and a more complete
answer would be j = 6, i = 4, and this is why:
i = 2;
j = i++ + ++i;
Obviously, the j line is more interesting, so
we'll talk about sequence of operations. The
preincrement operator is done before everything
else, so "++i" is done, giving i the value of
3, even before the addition, so you end up
with "j = 3++ + 3" which equals 6. After the
line is completed, i is incremented, causing
i to be 3++ = 4.
~Patrick
> From: James
> On Wed, 28 Apr 1999, Glynn Clements wrote:
>
> # > In gcc,
> # >
> # > if i = 2;
> # > then j = i++ + ++i;
> # >
> # > what is the value of j.
> #
> # i++ == 2
> # ++i == 3
> # => j == 5
>
> no. it's not. I typed this in:
>
> main() { int i = 2; printf ("%d", i++ + ++i);}
> and when i ran it i got 6 printed...
>
