Thread: what will be the output and why?

static variables retain their value throughout function calls. That is, if the function exits with staticvar = 7, the next call will start with staticvar = 7, etc.

it is also giving me same warning 2 times that function should return a value in line no 9 and 23.

Code:

#include <stdio.h>  /* for printf() etc */

int main()  /* main() returns an int; implicit main() is K&R */
{
    int a,b;
    a=sumdig(123);
    b=sumdig(123);
    printf("%d%d",a,b)
    return 0;  /* since main() is int, return a value (0 indicates success) */
}

int sumdig(int n)  /* again, implicit int rule is pre-ANSI */
{
    static int s=0;
    int d;
    if(n!=0)
    {
        d=n%10;
        n=(n-d)/10;
        s=s+d;
        return sumdig(n);  /* otherwise a garbage value is returned */
    }
    else return(s);
}

I changed the indentation too.

If you just want to understand how static local variables work, try this program:

Code:

#include <stdio.h>

void static_func(void);
void func(void);

int main(void) {
    int x;

    for(x = 0; x < 5; x ++) {  /* call static_func() and func() five times */
        static_func();
        func();
    }

    return 0;
}

void static_func(void) {
    static int x = 1;

    printf("%i\n", x++);
}

void func(void) {
    int x = 1;

    printf("%i\n", x++);
}

Note: static global variables are different. They're just like regular global variables but are only visible in the file in which they are declared. (You can have static functions, too, which act the same way -- one file only.)

Is this program the one that outputs "6 12"?

Code:

#include <stdio.h>

int main()
{
    int a,b;
    a=sumdig(123);
    b=sumdig(123);
    printf("%d%d",a,b)
    return 0;
}

int sumdig(int n)
{
    static int s=0;
    int d;
    if(n!=0)
    {
        d=n%10;
        n=(n-d)/10;
        s=s+d;
        return sumdig(n);
    }
    else return(s);
}

Okay, here's the first call:

Code:

sumdig(123)
d = n % 10;    d = 123 % 10;    d = 3;
n = (n-d)/10;    n = (123-3)/10;    n = 120/10;    n = 12;
s = s + d;    s = 0 + 3;    s = 3;
sumdig(12)
d = n % 10;    d = 12 % 10;    d = 2;
n = (n-d)/10;    n = (12-2)/10;    n = 10/10;    n = 1;
s = s + d;    s = 3 + 2;    s = 5;
sumdig(1)
d = n % 10;    d = 1 % 10;    d = 1;
n = (n-d)/10;    n = (1-1)/10;    n = 0/10;    n = 0;
s = s + d;    s = 5 + 1;    s = 6;

The second call is messed up because s isn't zero.

1)why are we taking here sumdig(123),sumdig(12) and sumdig(1)..from where can i come to know in this program that it would have to execute in the order of 123,then 12 and then 1.?....

It's a recursive function. If you mean how does the computer keep track of three different calls to the function, all the local variables for the function call are stored on the stack. Each call has its own distinct d and n (but not s, because s is stored on the heap).

2)...when would i know that i have to take the value of 6 and 12 amongst so many values from there...which we have computed and 6 is the value of which variable and 12 is the vaue of which variable?...didn't we have to print the values of ,,,a and b..?

Pardon? Do you mean that you're getting confused with so many variables involved?

3)which one of the values amongst 6 and 12 is of a and b?

Well, a is 6 and b is 12. See, b is 6 + whatever-was-in-s (which is 6).

basically how would the control flow execution would happen in the program..?

I tried to explain it in my previous post. If that doesn't help, I can try again. Or you could try a debugger.

You shouldn't take on two things at once, recursive functions and static local variables. Try examining my program above.