helllllp!!!!

Ok here is the source for a function which according to my book should raise x to the y power.
#include <stdio.h>



double pow(double , double );
int main()
{
double x, y, count, power;
for (count=1; count<=15; ++count)
{
printf("enter the number you would like to use\n", x);
scanf("%lf", &x);
printf("enter the power the number is placed to\n", y);
scanf("%lf", &y);
power = pow( x , y);
printf("the answer is %5.3lf\n", power);

return 0;
}

double pow(double x, double y)
{

return(power);
}

I keep getting 2 error messages the first one says 'pow' local functions not supported the other says unexpected end of file found. Now can anyone tell me where I've went wrong

Well firstly, you must be reading a CRAP book. Also, you're missing a closing brace for your for loop. As for that pow() function, well err.. yeah, that's just plain wrong. Get another book or read a few basic tutorials on functions in C.

Thats what I thought dude this book has been spitting one piece of bogus code right after another. The auther didn't eve have the sense to check his own code b4 he put it in the book

Well First off you might want to use another subject line like
"Im having problems using my pow function"

also code tags help. they are [ code] [ /code]


here is a quick fix that I can come up with

Code:

---

#include <stdio.h>
#include <math.h>

int main(void)
{
    double x = 0, y =0;
    printf("Enter x:");
    fflush(stdout);
    x = atoi(getch());
    printf("Enter y:");
    fflush(stdout);
    y = atoi(getch());
    printf("x^y = %f",pow(x,y);
    return 0;
}

---

You are trying to use inline functions..
I just looked over you code closer and you forgot to close your for loop..

Here is the fix to your code

Code:

---

#include <stdio.h>



double pow(double , double );
int main()
{
double x, y, count, power;
for (count=1; count<=15; ++count)
{
printf("enter the number you would like to use\n", x);
scanf("%lf", &x);
printf("enter the power the number is placed to\n", y);
scanf("%lf", &y);
power = pow( x , y);
printf("the answer is %5.3lf\n", power);
}
return 0;
}

double pow(double x, double y)
{
double power= x*y;
return(power);
}

---

:edit:
Blast Dog beat me to it while I was writing my reply.

If you want to learn C I would recomend that you get the K&R

Funny thing is that you both has printf() functions taking an argument but have no conversion specifier to take that argument.

Quote:

---

printf("enter the number you would like to use\n", x);
printf("enter the power the number is placed to\n", y);

---

You dont need the x, or the y.

I understand everything except this portion of the code

double pow(double x, double y)
{
double power= x*y;
return(power);
}

why x * y;
doesn't that say multiply x times y.

Quote:

---

Originally posted by drdodirty2002
I understand everything except this portion of the code

double pow(double x, double y)
{
double power= x*y;
return(power);
}

why x * y;
doesn't that say multiply x times y.

---

Yea.. I was writing it while helping someone reset there password. Sorry.

Sumpton I didnt even look at his printf() statements. I just fixed the for loop (you know cut and paste ;)

Quote:

---

Sumpton I didnt even look at his printf() statements. I just fixed the for loop (you know cut and paste ;)

---

I figured, and my name is stumon, lol. No biggy. Dr Dirty, something like this would pow the number.

Code:

---

double pow(double x, double y)
{
int i;
double x2 = x;
for (i = 0; i < y; i++){
x = x * x2;
}
return(x);
}

---

Or, you could include the math.h header and it has a pow() function in it. If you needed to use it.

LMAO.. Sorry about the name.. Doing Tech Support over the phone and tyring to respond at the same time can mess up my spelling. Thank god its friday!!!

Whadda bout this?

Code:

---

#include <stdio.h>

long getPow(long base, long power)
{
  if(power == 1)
      return base;
  else
      return base * getPow(base, power - 1);
}

int main()
{
  long num1, num2;
 
  printf("Enter the base: ");
  scanf("%ld", &num1);
 
  printf("Enter the power: ");
  scanf("%ld", &num2);
 
  printf("%ld^%ld = %ld", num1, num2, getPow(num1, num2));
 
  return 0;
}

---