Thread: preprocessor output

After preprocessing, your program becomes:

Code:

int main()
{
   int a,b=3;
   a=(b++*b++*b++);
   printf("%d %d",a,b);

}

Macros are not functions; they do dumb text replacement.

EDIT:

Originally Posted by patil.vishwa

Value for a is calculated as below with the b value as 3, since its post increment operator.

Actually, the expression results in undefined behaviour since b is read and changed multiple times within the same sequence point.

>> and then b is incremented 3 times. i.e, 4,5,6

Actually, if it worked as you thought it did (see laserlight's post), then the result would be,
3 * 4 * 5 = 60

Since it's post-increment, ie use the value and then increment. Still, not sure how you got 27 either, 4 * 5 * 6 = 120.

Such that,

Code:

int b = 3;
int stepOne = b++;   /* = 3, THEN b is incremented, such that b = 4 */
int stepTwo = b++;   /* = 4, THEN b is incremented, such that b = 5 */
int stepThree = b++; /* = 5, THEN b is incremented, such that b = 6 */

printf("%d,%d\n", (stepOne * stepTwo * stepThree), b);

Would output 60,6. Which is well defined behaviour, with several sequence points (one for each step).

Originally Posted by rohit_second

But the results are strange can anyone please explain me how the program behaves ????

Aside from the fact that b++ * b++ * b++ is undefined, you can actually find out what your code looks like AFTER the expansion of macros and inclusion of include files - it is sometimes quite a useful thing to do. Most compilers use an option -E or /E to output the preprocessed source code - usually you will need a -o or /o to tell the compiler where you want to output this file (it is commonly called .i, so "gcc -E myprog.c -o myprog.i" would be how you do this in gcc, for example).

Even if it doesn't actually answer your question here (because the actual answer has to do with undefined behaviour, and that is always impossible to answer categorically), it will be good practice to understand how to read pre-processed C or C++ code. Sometimes when you have a "missing semicolon" or some such, it is caused by a macro expansion that you didn't fully understand what it did.

--
Mats

Originally Posted by matsp

Aside from the fact that b++ * b++ * b++ is undefined, you can actually find out what your code looks like AFTER the expansion of macros and inclusion of include files.
--
Mats

I'd done this already, see it looks as i mentioned earlier...

Code:

int main()
{
   int a,b=3;
   a= ( ( b ++ ) * ( b ++ ) * ( b ++ ) );
   printf("%d %d \n",a,b);

}

The same program produces wierd result when pre increment operator is used.

Originally Posted by patil.vishwa

I'd done this already, see it looks as i mentioned earlier...

Code:

int main()
{
   int a,b=3;
   a= ( ( b ++ ) * ( b ++ ) * ( b ++ ) );
   printf("%d %d \n",a,b);

}

The same program produces wierd result when pre increment operator is used.

Yes, using ++, -- (and +=, -= and such) more than once on the same variable in the same expression will almost always cause "weird" output. This is because the C standard actually specifically says that updates to a variable should only ever be done ONCE between two sequence points [sequence points are places where calculations are completed, so usually the same as a statement or within the () of a function argument list]. Anything that does otherwise is "undefined behaviour", which essentially means that it can not be guaranteed in any way - it may produce the correct result, something "nearly correct", it may also produce completely silly results (e.g. you expect a value around 50, but it produces 11290823 because of some side-effect of one of the calculations), or even crash.

--
Mats