CODEWITHSUNDEEP
c
MNY
MNY

Reputation: 1536

Returning negative power on a double in C

I have a function that calculate the power of a double number, and the next task was to add the option so it could deal with negative power.

So I added this code to the function:

if (p < 0)
    {
        for (y = 1; y <= p; y++)
        {
            pow *= n;
        }
        pow = 1/pow;
    }

The whole program is short so I will share it too:

#include <stdio.h>
double power(double n, int p); // ANSI prototype

int main(void)

{
    double x, xpow;
    signed int exp;
    printf("Enter a number and the positive integer power");
    printf(" to which\nthe number will be raised. Enter q");
    printf(" to quit.\n");
    while (scanf("%lf%d", &x, &exp) == 2)
    {
        xpow = power(x,exp); // function call
        printf("%.3g to the power %d is %.5g\n", x, exp, xpow);
        printf("Enter next pair of numbers or q to quit.\n");
    }
    printf("Hope you enjoyed this power trip -- bye!\n");
    return 0;
}
double power(double n, signed int p) // function definition
{
    double pow = 1;
    int i;
    int y;

    if (p < 0)
    {
        for (y = 1; y <= p; y++)
        {
            pow *= n;
        }
        pow = 1/pow;
    }


    for (i = 1; i <= p; i++)
        pow *= n;

    return pow; // return the value of pow
}

If im entering the input 5.0 and the power -3 i suppose to get 0.008, and im getting 1...

Upvotes: 0

Views: 323

Answers (2)

Falmarri
Falmarri

Reputation: 48587

If you're entering your power as -3, none of your for conditions will be true.

   if (p < 0)
    {
        for (y = 1; y <= p; y++)
        {
            pow *= n;
        }
        pow = 1/pow;
    }

If p < 0, y will never be less than p.

Upvotes: 0

qPCR4vir
qPCR4vir

Reputation: 3571

Did you forget an else ? for the if

if (p < 0)
{
    for (y = 1; y <= -p; y++)
    {
        pow *= n;
    }
    pow = 1/pow;
}
else

for (i = 1; i <= p; i++)
    pow *= n;

Upvotes: 1

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