hello,
I am confused with following programs snippet code about how the
answer comes? statement/expression is evaluated?
1) int i=5;
i=i++*i++*i++*i++;
why the answer is not 5*6*7*8= 1680 but its 629?
2) int val=2;
val = - --val- val--- --val;
how to evaluate this?
3) int i=5;
printf("%d %d %d %d %d\n",++i,i++,i++,i++,++i);
why ans is not 6 6 7 7 10 but its 7 6 6 6 6?
regards,
rahul.
I should just direct you to the FAQ but I'm feeling generous today.
The short answer is that the result of any expression of the form
i=i++, i++*i++, etc., is undefined.
The long answer involves explaining how the autoincrement/decrement
operators actually work. The expression "i++" evaluates to the current
value of i; as a side affect, i is incremented by one, but exactly
*when* the side affect is applied isn't specified other than it must
happen before the next sequence point. A sequence point may be the end
of a statement, one of the Boolean operators && or ||, and (I think) a
comma operator (which is not the same thing as a comma separator in an
argument list).
So, given the statement
i = j++ * k++;
i is assigned the *current* value of j multiplied by the *current*
value of k; and sometime before the next sequence point, j and k are
incremented. The increments may happen as each expression is evaluted,
or after both j and k are evaluated but before the result is assigned
to i, or after the assignment to i. Each of the following are valid:
1. Update each object as it is evaluated; this is how most people
expect autoincrement/decrement to work, and it may actually work this
way occasionally:
t1 <- j
j <- j + 1
t2 <- k
k <- k + 1
i <- t1 * t2
2. Update objects after all expressions have been evaluated but before
assignment:
t1 <- j
t2 <- k
j <- j + 1
k <- k + 1
i <- t1 * t2
3. Update objects after all expressions have been evaluated and the
assignment has been carried out:
i <- j * k
j <- j + 1
k <- k + 1
The exact order depends on the compiler, the optimization settings, the
surrounding code, etc.
So basically, none of the expressions you provide should be expected to
yield correct results.
