WHAT IS THE ERROR IN THIS PROGRAM.(plz dont suggest any new program,only modificatio)

Code:

---

/*PROGRAM TO PRINT ARMSTRONG NUMBERS FROM 1 TO 1500*/
#include<stdio.h>
#define boundary 1500

*FUNCTION TO FIND X^N*/

int raise(int x,int n)
{if(n==1)
return x;
return x*raise(x,n-1);
}
/*FUNCTION TO ADD DIGITS*/

long int sum(int n,int noofdigit)
{        int temp;
        static long int sum1=0;
       
        sum1+=raise(n%10,noofdigit);
        if(n/10!=0)
        sum(n/10,noofdigit);
       
        else return sum1;
}
/*FUNCTION TO FIND NUMBER OF DIGITS*/
int digitn(int n)
        {int i;
        for(i=1;n/10!=0;++i)
                n/=10;
        return i;
        }
/*FUNCTION FOR CHECKING WHETHER NUMBER IS ARMSTRONG*/
void armstrong()
{
        int i,a,digits;
        for(i=0;i<=boundary;++i)
               
                if(sum(i,digitn(i))==i)
                printf("%d ",i);
               
       
        printf("\n");

}
main()
{
printf("ARMSTRONG NUMBERS IN FIRST 1500 NUMBERS ARE: ");
armstrong();


}

---

/*why i'm getting only 1 and 0?
I'm working in linux not TC
*/

Do you get any error/warning messages when you compile?

If so please post the complete error message.

Have you tried to print out intermediate results?


Jim

Hello JIM
no errors,no warnings
I've checked all functions.All working properly
Loop also works.

But what happens to printf() ?

Learn what "\n" means; add it to all of your printf, Or use flush stdout. This might not be problem, but not sure it is not a problem.

Learn to indent correctly.
Learn to comment your code.
Learn to use "{" and "}" even when they are not needed in loops.
Learn to test each function using test code.

Tim S.

Hello TIM
"\n" is of no importance in this program .just for printing another line.note that for loop has only one statement which is a compound statement

i removed "\n" ,the same problem...........................

HELLO TIM
See I've added comments
Each fucnctions are checked for different inputs

So you have tried something like:

Code:

---

int raise(int x,int n)
{
  if(n==1)
      return x;

  printf("%d %d\n", x, n);

  return x*raise(x,n-1);
}

---

I'm not saying that this is where the error is by the way.

Jim

Ya I've tried that test..............

For starters:

• Indent your code properly.
• Remove unused variables.
• Fully capitalise your macro names.
• When your function has a return type other than void, return something. (This is optional for main with respect to C99.)

I would expect something like this:

Code:

---

/*PROGRAM TO PRINT ARMSTRONG NUMBERS FROM 1 TO 1500*/

#include<stdio.h>

#define BOUNDARY 1500

/*FUNCTION TO FIND X^N*/

int raise(int x, int n)
{
    if (n == 1)
        return x;
    return x * raise(x, n - 1);
}

/*FUNCTION TO ADD DIGITS*/

long int sum(int n, int noofdigit)
{
    static long int sum1 = 0;

    sum1 += raise(n % 10, noofdigit);
    if (n / 10 != 0)
        sum(n / 10, noofdigit);
    else return sum1;
}

/*FUNCTION TO FIND NUMBER OF DIGITS*/
int digitn(int n)
{
    int i;
    for (i = 1; n / 10 != 0; ++i)
        n /= 10;
    return i;
}

/*FUNCTION FOR CHECKING WHETHER NUMBER IS ARMSTRONG*/
void armstrong()
{
    int i;
    for (i = 0; i <= BOUNDARY; ++i)
        if (sum(i, digitn(i)) == i)
            printf("%d ", i);
    printf("\n");
}

int main(void)
{
    printf("ARMSTRONG NUMBERS IN FIRST 1500 NUMBERS ARE: ");
    armstrong();
    return 0;
}

---

Now, compile your code with warnings turned on. You should get a warning about the sum function not returning a value for a control path. This is a problem. In fact, you should turn that static local variable in sum into a non-static local variable, and then use iteration, not recursion. Likewise, use iteration for the raise function (unless you want to use a more efficient implementation, in which case recursion might be an easy way out).

I added this to main

Code:

---

printf("Number of digits in 153: %d\n", digitn(153));
printf("Sum of digits in 153: %d\n", sum(153,digitn(153)));
printf("Number of digits in 10: %d\n", digitn(10));
printf("Sum of digits in 10: %d\n", sum(10,digitn(10)));

---

And, I got this

Code:

---

Number of digits in 153: 3
Sum of digits in 153: 153
Number of digits in 10: 2
Sum of digits in 10: 154

---

Do you see the problem?
See laserlight post for hint on fix.

Tim S.

TIM actually the sum() returns the sum as like in armstrong numbers.So sum(153,3) will return 1^3+5^3+3^3=153
153 is an armstrong number

Incase ofsum(10,digitn(10)); I got 1 which is 1^2 +0^2
I don,t know why you got 154

Thanks to laserlight for suggestions
But why should'nt recursion is not so good? Is it due to something like processing speed?................

Quote:

---

Originally Posted by linuxlover

But why should'nt recursion is not so good? Is it due to something like processing speed?

---

To perform recursion, the state of the function is saved. This can potentially use up the available stack space. There is also a limit to the recursive depth. These are not likely to be problems for your particular program, but when it is easy to use iteration instead of recursion, you might as well do so.

/*PROGRAM TO PRINT ARMSTRONG NUMBERS FROM 1 TO 1500*/

#include<stdio.h>

#define BOUNDARY 1500

/*FUNCTION TO FIND X^N*/

int raise(int x , int n)
{
int i,temp=x;

for(i=1;i<n;++i)
x*=temp;

return x;
}

/*FUNCTION TO ADD DIGITS*/
long int sum(int n,int noofdigit)
{ int temp;
long int sum1=0;

for(;n!=0;n/=10)
sum1+=raise(n%10,noofdigit);

return sum1;
}


/*FUNCTION TO FIND NUMBER OF DIGITS*/
int digitn(int n)
{
int i;
for (i = 1; n / 10 != 0; ++i)
n /= 10;
return i;
}

/*FUNCTION FOR CHECKING WHETHER NUMBER IS ARMSTRONG*/
void armstrong()
{
int i;
for (i = 0; i <= BOUNDARY; ++i)
if (sum(i, digitn(i)) == i)
printf("%d ", i);
printf("\n");
}

int main(void)
{
printf("ARMSTRONG NUMBERS IN FIRST 1500 NUMBERS ARE: ");
armstrong();
return 0;
}
/*THANKS LASERLIGHT THE ABOVE CODE WORKED*/

You're welcome :)

I'd be happier if you remembered to post in code tags though, heheh.