C Power function negative exponent without pow()

Question

I'm trying to make a little power calculator for learning purposes in C without using pow, but it always returns 0.00 when exponent is negative, please help.

full code:

#include<stdio.h>
//*  power caculator function

int power(x,y)
{
   float p=1.00;
   int i;
    if (y<0){
        y=-1*y;
        x=1/x;
    }
    for (i=1;i<=y;i++)
    {
        p=p*x;
    }

return p;
}



//*  main gets input, calls power caculator and prints result'
int main()
{
int b;
int e;
float p;
printf("enter base");
scanf("%d",&b);
printf("enter exponent");
scanf("%d",&e);
p=power(b,e);
printf("%d to the power of %d is %.2f",b,e,p);
return 0;
}
//* I am NOOB

Solution

You are using integers to hold decimal values, in this case with x and with the return type of the power function.

try:

float power(x,y)
{
   float p=1.00;
   float xx = (float)x;
   int i;
    if (y<0){
        y=-1*y;
        xx=1/xx;
    }
    for (i=1;i<=y;i++)
    {
        p=p*xx;
    }

return p;
}

Comments

[Weather Vane]

int power(x,y) what are the argument types? Also please pay attention to compiler warnings. By omitting the type, then compiler assumes they are int. You did supply int arguments but x=1/x; assumes integer division.

[Christian Gibbons]

What is 1/x for any x > 1?

[Lee Daniel Crocker]

pow() works just fine with negative exponents, but then you chop them off by declaring your function to return int.

[infinite]

Truncation is a problem apart from the other basic ones. 1 divided by any int is 0.

[Weather Vane]

The return value of power() is int. So even when you get the argument types (or the casting) right, any negative power (result < 1) will return 0.

[Christian Gibbons]

@infinite almost, but not quite. 1 and -1 will give you a different result. And 0 makes babies cry.

[infinite]

@ChristianGibbons I tried 1/i, 1/(-i), -1/(i), -1/(-i) where i is an integer other than 0 and got the same result 0 on gcc 6.3

[Christian Gibbons]

@infinite 1 is an integer other than 0. I would be very interested in a system where 1/1 comes out to 0.

[NOOBAF]

Ah I SEE NOW! Thanks Miguel!

[Miguel Mota]

Great! Best regards!

[Christian Gibbons]

Not the downvoter, but I assume it's because the function still has not explicitly defined the types of x and y.

[Miguel Mota]

yea... defining types for parameters will help a lot... I agree.. I'll down vote myself... jejeje