operator confusion

[jam7cacci]

I'm trying to add all the square root answer but why am i getting this output: -858993430

#include <iostream.h>


void main()
{

    int input;
    int sqrt;
    int i, sum, add;

    cout << "Please enter your desire number: ";
    cin  >> input;

    for (i=1; i <= input; i++)
    {
        sqrt = i * i;       
        add = sqrt;
        cout << i << "^2 = " << sqrt<< "\n";
        sum +=  add;
    }   

    cout << "the sum is: " << sum;

}

what could be wrong? hope you could help me with this. thanks!

[nucleon]

sum must be initialized to zero.
You don't need the variable add.

[Clockowl]

And, not related to programming, you're squaring it, not square rooting it.

[siddhant3s]

Please. Please
Don't use void main. http://cppdb.blogspot.com/2009/02/should-i-use-void-main-or-int-main-or.html
And yes, you are adding the squares of the integer rather than the square root.
To find the square root, you need to #include <cmath> and use the pow() function.http://www.cppreference.com/wiki/c/math/pow

[Clockowl]

To find the square root, you need to #include <cmath> and use the pow() function.http://www.cppreference.com/wiki/c/math/pow

He means the sqrt() function.

[siddhant3s]

No, I really meant pow() function.
It is more versatile than sqrt().
You can use pow(N,0.5) rather than sqtr(N)

[Clockowl]

xD Okay, sorry. I'd suggest sqrt() nonetheless, but it doesn't really matter.

[jam7cacci]

Please. Please
Don't use void main. http://cppdb.blogspot.com/2009/02/sh...t-main-or.html
And yes, you are adding the squares of the integer rather than the square root.
To find the square root, you need to #include <cmath> and use the pow() function. http://www.cppreference.com/wiki/c/math/pow

thanks!

i got it! hmmm.... i cant use other than void... i'm not a pro yet and i must stick with the way my prof has taught us. though yah it would be nice to experiment from time to time but as now i have to stick with it :(. new ways are hard for me to understand especially when you have tones of assignments regarding it! hahaha!

but anyway, i appreciate it! thank a lot everybody!

so here's my code and glad to have it work... after how many paper scratches for computation.. :(

#include <iostream.h>


void main()
{

    int input;
    int sqrt;
    int i, sum, add;


    cout << "Please enter your desire number: ";
    cin  >> input;


    cout << "\n" ;

    for (i=1; i <= input; i++)
    {

        sqrt = i * i;       

        cout << "\n" << i << "^2 = " << sqrt<< "\n";

        sum = sum + sqrt;



    }



    add = 0;

    for (i=0; i <=input; i++)
    {
        sum = i * i;
        add = add + sum;
    }   


    cout << "\nThe sum is: " << add << "\n\n";
}

now my next problem is how to do the sum of powers from 1 to n :(

when the user inputted 4

it should be

1^1 = 1
2^2 = 4
3^3 = 27
4^4 = 256

whew... :(

now it's getting more confusing....

[jam7cacci]

No, I really meant pow() function.
It is more versatile than sqrt().
You can use pow(N,0.5) rather than sqtr(N)

hahaha!

i tried the pow! ya you're right it was much easier to use...

thanks again!