Understanding Pre and Post Incrementation in C

[the_kool_guy]

Homework Statement

initially i = 10;

Homework Equations

1) x = i++ + i++
2) x = ++i + ++i
3) x = i++ + ++i
4) x = ++i + i++
5) x= i++ + ++i + i
6) x= ++i + i++ + i

The Attempt at a Solution

i++ is post incrementation while ++i is pre incrementation
by this logic i know ans of eqn 1) is 20.but other equations gets me wrong
by gcc, answers to 2),3),4),5),6) are 24,22,22,33,33 .
i am getting no idea how this thing works.
somehow i am able to think of a way to answer above but situation gets worse in case of
x = i++ + ++i + i++ + ++i;
and x= ++i + i++ + ++i + i++;
where answers are 45 and 46.
this whole thing is screwing me up
pls help
thanks

 

[TylerH]

Which has precedence, preincrement or +? Preinc does.

i = 10
2)++i + ++i == 12 + 12 because the preincrement is done BEFORE the addition, AND it is does TWICE.

Sometimes it helps to break down compound expressions into separate sequential lines in order of their precedence.
++i + ++i becomes
++i;
++i;
i + i;
The ++i's come first b/c they are higher in precedence.

 

[I like Serena]

I've got some bad news for you.

According to the C/C++ standard, the value of all of these expressions is undefined.
That is, each compiler implementation might give a different result.
Furthermore, depending on compiler flags (typically optimization), the results might come out different as well.

At least in the current implementation of gcc, the results are the same independent of the optimization flags, but in previous versions this was not so.

Cheers!

 

[D H]

According to the C/C++ standard, the value of all of these expressions is undefined.

Bingo.

The results of each and every one of these expressions is undefined. Each expression is modifying an object, the variable i in this case, more than once between sequence points. That is quintessential undefined behavior.

 

[the_kool_guy]

Which has precedence, preincrement or +? Preinc does.

i = 10
2)++i + ++i == 12 + 12 because the preincrement is done BEFORE the addition, AND it is does TWICE.

Sometimes it helps to break down compound expressions into separate sequential lines in order of their precedence.
++i + ++i becomes
++i;
++i;
i + i;
The ++i's come first b/c they are higher in precedence.

this way answer comes in multiple of number of terms.
what about the value of x in which answer is 45 and 46...

 

[the_kool_guy]

Bingo.

The results of each and every one of these expressions is undefined. Each expression is modifying an object, the variable i in this case, more than once between sequence points. That is quintessential undefined behavior.

that means there is no logic to define these behavior?
but this is surely got to come in my exams... :(

 

[the_kool_guy]

I've got some bad news for you.

According to the C/C++ standard, the value of all of these expressions is undefined.
That is, each compiler implementation might give a different result.
Furthermore, depending on compiler flags (typically optimization), the results might come out different as well.

At least in the current implementation of gcc, the results are the same independent of the optimization flags, but in previous versions this was not so.

Cheers!

The results of each and every one of these expressions is undefined. Each expression is modifying an object, the variable i in this case, more than once between sequence points. That is quintessential undefined behavior.[/QUOTE]

that means there is no logic to define these behavior?
but this is surely got to come in my exams... :(

 

[I like Serena]

that means there is no logic to define these behavior?
but this is surely got to come in my exams... :(

If it does, choose the answer that says: the result is undefined.
Or perhaps the one that says: the result is implementation defined.

And if you want to practice your skills at these operators, try this one:

What does the following do?

while (*d++ = *s++);

where s and d are char pointers.

Or this one?

for (unsigned i = 5; i-- > 0;)
{
    printf("%d", i);
}

 

[TylerH]

Who looks like an idiot? I do!

Yet another good reason to use -Wall.