从函数返回浮点值

Question

hey i really need help with this program.. i doesnt want return a proper answer for a negative power always return 1.. can anyone help?? The program should help the user to enter a base number and power and when executed calculates the value and displays the result

float square (float a,int b);

int main(){
    int power;
    float base;

    printf("Please input your base number :\n");
    scanf("%f",&base);
    printf("Please input your power :\n");
    scanf("%d",&power);

    float answer = square(base,power);

    printf("Your Result is :\n%f\n",answer);

    system("pause");
    return 0;
}

 float square (float a,int b){
     int counter;
     float total = 1;
     if (b==0){
         return 1;
     } else if (b>0){
         for(counter=0;counter<b;counter++){
             total = total*a;
         }
         return total;
     } else {
         for (counter=0;counter<b;counter++){
             total = total*a;
         }
         total = 1 / total;
         return total;
     }
}

Answer 1

Declare total as a float instead of an int .

Also, your if's should be on b instead of a .

After that, remove all those unnecessary (float) castings.

Also, your last for for should run backwards as your initial value is less than 0 .

Answer 2

Your square function should look like this:

float square (float a,int b){
   int counter;
   float total = 1;
   if (b==0){
       return 1;
   } else if (b>0){
       for(counter=0;counter<b;counter++){
           total = (float)total*a;
       }
       return total;
   } else {
       for (counter=0;counter< -b;counter++){
           total = (float)total*a;
       }
       total = 1.0 / (float)total;
       return total;
   }
}

You need a float for total, you need to check b instead of a in your if statements, and you need to count up to -b when it's negative.

Answer 3

Declare total to float and initialize it as

float total = 1.0f;   

Use abs function for dealing with -ve b .

else {
         for (counter=0;counter < abs(b);counter++){
             total = total*a;
     }
     total = 1 / total;
     return total;

}
and include <stdlib.h> .
See the working code Here .

Answer 4

This part

 } else {
     for (counter=0;counter<b;counter++){
         total = total*a;
     }
     total = 1 / total;
     return total;
 }

ignores the fact that b is < 0.

So either you should put b = -b; in, or

     for (counter=0;counter>b;counter--){
         total = total / a;
     }
     return total;

might help you.

Answer 5

This does the job ;)

#include <stdio.h>

#include <math.h>


int main(){
   int power;
   float base;

   printf("Please input your base number :\n");
   scanf("%f",&base);
   printf("Please input your power :\n");
   scanf("%d",&power);

   float answer = pow(base,power);

   printf("Your Result is :\n%f\n",answer);

   system("pause");
   return 0;
}

Answer 6

Here's a cool trick:

y = a^b

ln(y) = b ln(a)

y = exp(b ln(a))

so the function should look like this:

double Power(double a, double b)
{
    return exp(b*log(a));
}