the value of a is 11, how??

#define SQR(x) (x*x)
int main()
{
	int a,b=3;
	a=SQR(b+2);
	printf("a=%d\n",a);
	return 0;
}

Meenu a �crit :
> the value of a is 11, how??
> 
> #define SQR(x) (x*x)
> int main()
> {
> 	int a,b=3;
> 	a=SQR(b+2);
> 	printf("a=%d\n",a);
> 	return 0;
> }
> 
a = SQR(b+2) where SQR --> "b+2"
a = b+2*b+2
a = 3+(2*3)+2
a = 3+6+2
a = 11

On 2005-07-27, Meenu wrote:
> the value of a is 11, how??
>
> #define SQR(x) (x*x)

#define SQR(x) ((x)*(x))

    Otherwise:

3 + 2 * 3 + 2 = 3 + (2 * 3) + 2 = 11

> int main()
> {
> 	int a,b=3;
> 	a=SQR(b+2);
> 	printf("a=%d\n",a);
> 	return 0;
> }


-- 
    Chris F.A. Johnson                     <http://cfaj.freeshell.org>
    ==================================================================
    Shell Scripting Recipes: A Problem-Solution Approach, 2005, Apress
    <http://www.torfree.net/~chris/books/cfaj/ssr.html>

Meenu wrote:
> the value of a is 11, how??
> 
> #define SQR(x) (x*x)
> int main()
> {
> 	int a,b=3;
> 	a=SQR(b+2);

is
         a=b+2*b+2;
since the value of b is 3, the value of the expansion is
         a=3+6+2;
          = 11;
This is why, if you are _sure_ that there are no side effects in the 
arguments to SQR() it should be
  #define SQR(x) ((x)*(x))
But, of course, you have no such guarantees about no side effects.  Both 
versions die with
      SQR(++b);
for example.
So, what you really want is
inline int square_int(int x) { return x*x; }





> 	printf("a=%d\n",a);
> 	return 0;
> }
> 

hi meenu ,

see the problem here is that all the #defines are taken care at
preprocessing level .
So wat the compiler gets is not 5*5 but actually 3 + 2 * 3 + 2
which is 11


Meenu wrote:
> the value of a is 11, how??
>
> #define SQR(x) (x*x)
> int main()
> {
> 	int a,b=3;
> 	a=SQR(b+2);
> 	printf("a=%d\n",a);
> 	return 0;
> }

"abhik" <abhikbagchi@gmail.com> writes:
> Meenu wrote:
> > the value of a is 11, how??
> >
> > #define SQR(x) (x*x)
> > int main()
> > {
> > 	int a,b=3;
> > 	a=SQR(b+2);
> > 	printf("a=%d\n",a);
> > 	return 0;
> > }
>
> see the problem here is that all the #defines are taken care at
> preprocessing level .
> So wat the compiler gets is not 5*5 but actually 3 + 2 * 3 + 2
> which is 11

Which can easily be solved by proper use of parentheses:

        #define SQR(x) ((x) * (x))

This macro has yet another potential "bug", which is caused by using the
macro argument twice.  If the "x" expression has side-effects they will
be doubled, yielding strange results in invocations like:

        SQR(b++, b++)

But that's probably not what the original poster asked for :)

Giorgos Keramidas wrote:

>         #define SQR(x) ((x) * (x))
> 
> This macro has yet another potential "bug", which is caused by using the
> macro argument twice.  If the "x" expression has side-effects they will
> be doubled, yielding strange results in invocations like:
> 
>         SQR(b++, b++)

....which would most likely not compile :-)

(SQR takes ONE parameter)

Peter

Peter Pichler wrote:
> Giorgos Keramidas wrote:
> 
>>         #define SQR(x) ((x) * (x))
>>
>> This macro has yet another potential "bug", which is caused by using the
>> macro argument twice.  If the "x" expression has side-effects they will
>> be doubled, yielding strange results in invocations like:
>>
>>         SQR(b++, b++)
> 
> 
> ...which would most likely not compile :-)
> 
> (SQR takes ONE parameter)

Moreover, even if the function *did* take two parameters it would cause 
problems (undefined behavior) even if it was a "true" function 
(modification of a variable twice with no intervening sequence point).

August