Member Avatar for 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!

Edited by mike_2000_17 because: Fixed formatting
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As pointed out, the original negative result was classic "garbage" from using sum before initializing it. Local automatic variables are indeterminate; initializing accumulators (for example int sum = 0;) removes that class of bug. Also note 's point that the code was squaring, not square-rooting, and 's remark that int main() is the standard signature (many compilers accept void main, but it is non-standard).

To produce the sequence 1^1, 2^2, ..., n^n and their sum safely, integer exponentiation is preferable to std::pow (which returns double and can round or overflow). The following example uses exponentiation by squaring, checks for 64-bit overflow, prints each i^i, and accumulates a sum:

#include <iostream>
#include <cstdint>
#include <limits>

unsigned long long ipow(unsigned long long base, unsigned exp) {
    unsigned long long res = 1;
    while (exp) {
        if (exp & 1) {
            if (base && res > std::numeric_limits<unsigned long long>::max() / base) return 0;
            res *= base;
        }
        exp >>= 1;
        if (exp) {
            if (base && base > std::numeric_limits<unsigned long long>::max() / base) return 0;
            base *= base;
        }
    }
    return res;
}

int main() {
    unsigned n;
    if (!(std::cin >> n)) return 0;
    unsigned long long sum = 0;
    for (unsigned i = 1; i <= n; ++i) {
        unsigned long long p = ipow(i, i);
        if (p == 0) { std::cout << i << "^" << i << " overflows 64-bit\n"; break; }
        std::cout << i << "^" << i << " = " << p << '\n';
        if (sum > std::numeric_limits<unsigned long long>::max() - p) { std::cout << "sum would overflow\n"; break; }
        sum += p;
    }
    std::cout << "sum = " << sum << '\n';
    return 0;
}

Notes: unsigned long long still overflows quickly (e.g. large n); for arbitrarily large integer results use a multiprecision library. Avoid naming variables after standard functions (for example sqrt) and remove unused temporaries (like add) to keep code clear.

Recommended Answers

All 8 Replies

Member Avatar for nucleon

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

Member Avatar for Clockowl

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

Member Avatar for Clockowl

To find the square root, you need to #include <cmath> and use the pow() function.

He means the sqrt() function.

Member Avatar for siddhant3s

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

Member Avatar for Clockowl

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

Member Avatar for 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.

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....

Edited by mike_2000_17 because: Fixed formatting
Member Avatar for 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!

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