While loop for sum of cubes

**ralphwiggam** · 09-28-2006

I feel like I'm so close. I have to write a sum of cubes beginning with 1 to whatever # "n". I am restricted to a while loop. I just completed a similar sum of squares using a for loop and I got it fairly quickly.

Can anyone suggest something?

Code:

#include <stdio.h>
#include <math.h>

main ()
{
/* Declare variables */
  int i, n, sumcubes = 1;

/* Set for loop conditions */
  while (i <= n) 
  {
  sumcubes += (i*i*i);
  printf("Enter n: \n");
  scanf("%i", &n);

  printf ("The sum of the cubes from 1..%d is %d\n", n, 
      sumcubes); 

  }

 system("pause");
 exit(0); 

}

**twomers** · 09-28-2006

I think you should ask the user for the number of loops before the while() loop.

Also, i and n aren't set in the boundaries of the while() loop. This leads to undefined behaviour!

Code:

int i=0, n;

printf("Enter the loop#: ");
scanf("%d", &n);

while (i<=n){ ... }

Unless you're doing something different

**Driver** · 09-28-2006

In addition, you really ought to declare the main() function as returning an int.
See here for further info.

**ralphwiggam** · 09-28-2006

Thank you all for your suggestions.

twomers. . .are you saying I should set the i and n variables within the loop or out of it?
i =1 not zero at the beginning per the equation I have so I cannot set i=0.
I do not want to ask how many times the user wants to loop.
I want the program to increment by one from the #1 until the arbitrary # n and take all the cubes and add them up so once i increments by one until it reaches "n" it will turn false and exit the while loop.

Thank you too, Angoid.

This still does not work.

Code:

  while (i <= n) 
  {
  printf("Enter n: \n");
  scanf("%i", &n);
  ++i;
  sumcubes += (i*i*i);
  printf ("The sum of the cubes from 1..%d is %d\n", n, 
      sumcubes); 

  }

**LineOFire** · 09-28-2006

Originally Posted by ralphwiggam

Thank you all for your suggestions.

twomers. . .are you saying I should set the i and n variables within the loop or out of it?
i =1 not zero at the beginning per the equation I have so I cannot set i=0.
I do not want to ask how many times the user wants to loop.
I want the program to increment by one from the #1 until the arbitrary # n and take all the cubes and add them up so once i increments by one until it reaches "n" it will turn false and exit the while loop.

Thank you too, Angoid.

This still does not work.

Code:

  while (i <= n) 
  {
  printf("Enter n: \n");
  scanf("%i", &n);
  ++i;
  sumcubes += (i*i*i);
  printf ("The sum of the cubes from 1..%d is %d\n", n, 
      sumcubes); 

  }

I think he meant that you aren't initliazing i and n before the loop so those variables could contain anything. You never set what i is so it will not contain 1 like you want it to.

Can you post a sample of what you want the input/output of your program to be?

Is this what you want your program to do?:

Enter n:
1
The sum of the cubes from 1..1 is 1

Enter n:
2
The sum of the cubes from 1..2 is 9

Enter n:
3
The sum of the cubes from 1..3 is 36

Enter n:
4
The sum of the cubes from 1..4 is 100

**ssharish2005** · 09-28-2006

Code:

printf("Enter n: \n");
  scanf("%i", &n);
while (i <= n) 
  {
  sumcubes += (i*i*i);
  
  }
printf ("The sum of the cubes from 1..%d is %d\n", n, 
      sumcubes);

you are almost there. Just need to place your bit of code above while loop and a bit below the while loop

ssharish2005

**ralphwiggam** · 09-28-2006

LineOFire
That is exactly what I want my program to do.
I would put down exactly what u did but that is a waste of time, u nailed it.
When I set i down just as 1 it does not work as well as the code that I have below.

ssharish2005
I did exactly what u suggested but now it executes even worse than before. It only asks for "n" and does nothing with it.
Back when I had my "printf" and "scanf" functions within the while loop it at least took n and cubed the n and added 1 displaying the last printf I had in the loop with this code (ie this code gets me closer than the other variations I have to the output I want the sum of cubes over the range 1-"n"):

Code:

  int i, n, sumcubes = 1;

/* Set for loop conditions */
  while (i <= n) 
  {
  printf("Enter n: \n");
  scanf("%i", &n);
  ++i;
  sumcubes += (i*i*i);
  printf ("The sum of the cubes from 1..%d is %d\n", n, 
      sumcubes); 

  }

Thank you for your suggestions

PS I went to twomers site to thank him/her but my browser said it was down

**quzah** · 09-28-2006

You are first using n uninitialized. Read a value for it ahead of time; use a do-while loop, or initialize the variable first.

Quzah.

**ssharish2005** · 09-28-2006

Code:

printf("Enter a value for n\n?");
    scanf("%d",&n);
    
    while(i++<n)
        sum+=(i*i*i);
        
    printf("The total sum is %d\n",sum);

ssharish2005

**ralphwiggam** · 09-28-2006

ssharish2005 . . .I tried your latest suggestion but it asks for a number & does nothing with it.

Code:

  int i = 1, n = 1, sumcubes = 1;

  printf("Enter n: \n");
  scanf("%i", &n);
    while (i++<n) 
  {
  sumcubes += (i*i*i);
  printf ("The sum of the cubes from 1..%d is %d\n", n, 
      sumcubes); 
  }

Oops I accidentally hit the post reply button too early but I can tell you

quzah I tried it your way & I guess it's better now but still does not yield to my desire (no double meaning)

Code:

{
/* Declare variables */
  int i = 1, n = 1, sumcubes = 1;

/* Set for loop conditions */
  while (i <= n) 
  {
  printf("Enter n: \n");
  scanf("%i", &n);
  ++i;
  sumcubes += (i*i*i);
  printf ("The sum of the cubes from 1..%d is %d\n", n, 
      sumcubes); 
  }

**ssharish2005** · 09-28-2006

Code:

#include<stdio.h>

int main()
{
    int i=0,n;
    int sum=0;
    
    printf("Enter a value for n\n?");
    scanf("%d",&n);
    
    while(i++<n)
        sum+=(i*i*i);
        
    printf("The total sum is %d\n",sum);
    
    getchar();
    return 0;
}
/*my output
Enter a value for n
?3
The total sum is 36
*/

This is how u are suppose to use that bit of code. If it ask for some value enter some value see what sort of output it gives you. The value which it was asking you is for the variable n. Look at my code, u will know what it is , after calcualtion things it prints of the total sum. If u place the printf fucntion inside the while it will show you the each iteeration sum results. If u want to debug the program u can do that.

ssharish2005

**ralphwiggam** · 09-28-2006

ssharish2005
I cut and pasted the whole code & it executes and asks for n, but once I designate a # it closes down. I am using Bloodshed software C++ version 4.9.9.2 but I save everything as .c files before I compile and run them.
I am sorry to try your patience.

**quzah** · 09-28-2006

FAQ

Quzah.

**ralphwiggam** · 09-29-2006

ssharish2005
That was the key. It works! I really do appreciate it.
Thank you too quzah.
Thanks to everyone!
I am learning a lot, thanks.

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