Thread: help in find perfect numbers

Hmm, don't perfect numbers involve summing all the integer divisors? I.e. I think your if (sum == i) is probably in the wrong place. Edit: well, now that I look at it, it's not in the incorrect place (fix that formatting!) What is your loop doing? Calculating the aliquot sum of what (it looks like it's only going to do it for 1)? If so then everything looks right I guess -- to a point. Try adding a \n to your printf().

Edit:
Also:
a) You will benefit from indenting your code correctly;
b) void main() is not a valid declaration for main

Edit: You should reset sum to 0 before each iteration of the while loop

Why don't you then try starting from a piece of code that works, and work your way up from there?

... You tried for 4 hours?! Try more!!! Once, I pulled an all-nighter, 24+ hours trying to figure out why my code wouldn't work! Non-stop!! The only thing I'd eaten/drunk during that was coffee, water and 6 crackers!
( Well, ok, maybe it's just me, but I can't relax when I can't find the bug )

I haven't compiled your program.

How many times will the loop below be executed?

Code:

for(j=1;j<i;j++)

Given that 'j' is 1 and 'i' is also 1?

Edit: I'll give you a hint because you're trying (unlike some other people asking questions).

Your current code is, effectively, the same as:

Code:

#include<stdio.h>

int main(void)
{
    int i=1,n,sum=0,count=1,j;
    printf("HOw many no u want to print");
    scanf("%d", &n);
    while(count <= n)
    {
        i++;
        if (sum == i)
        {
            printf("%d",sum);
            count++;
        }
    }
    return 0;
}

Why would this be the case? When will the condition if (sum == i) evaluate to true?

Why does the while loop, apparently, never end (unless the number you enter for 'n' is 1 (and only 1)? And if this is the case, your code is, effectively, the same as:

Code:

#include<stdio.h>

int main(void)
{
    int i=1,n,sum=0,count=1,j;
    printf("HOw many no u want to print");
    scanf("%d",&n);
    return 0;
}