Model for exponent function

[baj6]

Having difficulties getting values for negative exponents. Keep getting 0. Any help would be appreciated.* side note: I know I can just use pow function w/ cmath but practicing use of for loops.*

#include <iostream>

using namespace std;

int exponentFcn (int baseNum, int powNum) {
    double result;
    int i;
    if (powNum >= 0){
        result = 1;
        for (int i = 0; i < powNum; i++){
        result = result * baseNum;
    }
    return result;
    }
    else if (powNum < 0){
        result = 1;
        for (int i = 0; i > powNum; i--){
        result = result * baseNum;
    }
    return 1 / result;
    }
    else {
        cout << "Invalid entry";
    }

}
int main()

{
    int baseNum;
    int powNum;
    cout << "Enter base number: ";
    cin >> baseNum;
    cout << "Enter power: ";
    cin >> powNum;
    cout << "Answer: " << exponentFcn(baseNum,powNum);
    return 0;
 }

[salem c]

> return 1 / result;
Any negative power will always be zero, in integer arithmetic.

> int exponentFcn (int baseNum, int powNum)
Try returning a double, like what the type of result is already.

[dutch]

What exactly do you think is happening here?

for (int i = 0; i > powNum; i--)

And your else clause will never happen. You should get rid of it and the else if (...) should just be an else.

[JLBorges]

#include <iostream>
#include <cmath>
#include <iomanip>

constexpr double ipow( int base, int exp )
{
    if( exp == 0 ) return 1 ; // note: ipow(0,0) == 1 (this mimics std::pow)
    else if( exp<0 ) return 1.0 / ipow( base, -exp ) ;
    else return base * ipow( base, exp-1 ) ;
}

int main()
{
    std::cout << std::fixed << std::showpos ;

    for( int base : { -2, -1, 0, 1, 2 } )
        for( int exp : { -5, -2, -1, 0, 1, 2, 5 } )
        {
            std::cout << base << " ^ " << exp
                      << " == " << std::setw(10) << ipow( base, exp )
                      << " == " << std::setw(10) << std::pow( base, exp ) << '\n' ;
        }
}

http://coliru.stacked-crooked.com/a/101370b37011b253

[baj6]

@salem c thanks. I just had to change the function and its parameters to doubles.

[jonnin]

another way... using an unrolled loop, any input type will work long as the exponent itself is only an int (double type exponents are tricky).

http://www.cplusplus.com/forum/beginner/250227/

if you want to support negative exponent, would need to 1/x the result at the end AND avoid tapping the sign bit as part of the power, similar to what you did but watch that sign bit in this approach.

[lastchance]

#include <iostream>
#include <iomanip>
#include <cmath>
using namespace std;

double ipow( int base, int exponent )
{
   if ( exponent < 0 ) return 1.0 / ipow( base, -exponent );

   double result = 1.0;
   for ( ; exponent; base *= base, exponent /= 2 ) if ( exponent % 2 ) result *= base;
   return result;
}


int main()
{
   #define FORMAT << fixed << showpos << setw( 12 ) <<

   for ( int base : { -5, -2, -1, 0, 1, 2, 5 } )
   {
      for ( int exponent : { -5, -2, -1, 0, 1, 2, 5 } )
      {
         cout << base << " ^ " << exponent 
              << " = " FORMAT ipow( base, exponent )
              << " = " FORMAT pow( base, exponent ) << '\n';
      }
   }
}