RE: [LIP] Doubt in ++ operator

> int i=7; int j=i++ * i++; printf("%d",j);
> The above set of instruction will give you 7 * 8 = 56.

To add to previous posts, consider this expression:
j = i++ * ++i;

In this expression, in addition to undefined behavior that
is invoked (refer Binand's and Raj's post), you cannot
predict absolutely *anything* The C language does *not*
define the order of evaluation of sub-expressions in an
expression - so here you don't even know which one will be
evaluated first, i++ or ++i.

Best thing, IMHO, stay *away* from such expressions, and
never, ever make a mistake of using them in production
code.

Here is the definition of Undefined Behavior, from the
standard:

undefined behavior
Behavior, upon use of a nonportable or erroneous program 
construct, of erroneous data, or of indeterminately valued 
objects, for which this International Standard imposes no
requirements.

Which roughly translates to means - your compiler is free
to do *anything* if undefined behavior is invoked, which
can range from terminating the program, to getting 1000
pizzas' delivered to your doorstep, to charging your credit
card a transaction of USD 100,000. ;)

Cheers,
Amar

--
Infinity is where God is dividing by zero.


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