C Programming - Declarations and Initializations - Discussion

Can you give me the brief explanation about linkages and what are the storage classes?

Thanx Dixin

The reason is
x=5
++x means x=6
then ++x means x=7
then only it will start to add
z=x+x
but at the end x=7
therefore
z=7+7
ie z=14 is the answer

x= 5;
z = ++x + ++x ;
cout<<z;

why does this display 14 , instead of 13 ?

What Prathyusha said is correct. That can be easily understood using this program.

#include<stdio.h>
int main()
{
printf("\n%d",funi());
printf("\n%d",funi());
printf("\n%d",funi());
printf("\n%d",funj());
printf("\n%d",funj());
printf("\n%d",funj());
//printf("\ni=%d\nj=%d",i,j);
return 0;
}

int funi()
{
int i;
i++;
return i;
}

int funj()
{
static int j;
j++;
return j;
}


output:
1
1
1
1
2
3

Good explanation pratyusha, thanq so much.

I will thank full to Satish Narasimman and Prathyusa, who cleared my doubt.

We declare already a static varible, in derived class we declare one more same static variable as declared first with different value, and call the previous variable,.......then.....Is the out put of program is first variable's value or secon variale's value...

//SATHISH explanation was correct!..
int main()
{
printf(" %d",root());
printf(" %d",root());
}

int root()
{
static int i;
i++;
return i;
}

BY DEFAULT ,ANY INTEGER OR STATIC INT VARIABLE VALUE IS ZERO.
BUT WHEN EACH TIME THE FUNCTION WAS CALLED ,THE DIFFERENCE OF NORMAL INT AND STATIC INT CAN BE SEEN.
IF THE "i" HERE IS NORMAL INTEGER IT WILL JUST RETURN ALWAYS VALUE "1" ,EVERY TIME IT IS CALLED .
BUT THE STATIC VARIABLE WILL HOLD PREVIOUS VALUES OF "i".EVERY TIME WHEN THE FUNCTION IS CALLED IT WILL THE PREVIOUS VALUES ARE RESTORED.

Can static variables be declared before main() ?