Trying to Approximate Euler's number in C - c

Question

I am trying to approximate Euler's number using the formula (1+(1/n))^n.
The compiler is telling me that there is an "expected expression before 'double'"
Here is the code:
#include <stdio.h>
#include <math.h>
int main()
{
int x, y, power;
int num = 1;
int position = 1;
while (position <= 100)
{
num = 1/num;
num = num + 1;
x = num;
power = double pow(x, x); //here
printf("%f", power);
position += 1;
num = position;
}
}

Answer 1

If you want a number to be a double (number with decimals), you need to define it as a double, not an integer. I have this code which should solve your problem. Also make sure to compile gcc FILEPATH -lm -o OUTPUTPATH if you are using UNIX.
#include <stdio.h>
#include <math.h>
int main()
{
double x, y, power, num = 1; //doubles allow for decimal places so declare it as double
int position = 1; //Position seems to only be an integer, so declare it as an int.
while (position <= 100)
{
num = 1/num;
num++;
x = num;
power = pow(x, x);
printf("%f", power);
position += 1;
num = position;
}
}
Another option is a for loop:
#include <stdio.h>
#include <math.h>
int main()
{
double x, y, power, num = 1;
for (int i = 1; i <= 100; i++) {
num = 1/num;
num = num + 1;
x = num;
power = pow(x, x);
printf("%f", power);
position += 1;
num = i;
}
}
If you are trying to approximate Euler's number, I don't see why not just try something like:
static const double E = 2.718281828459045;
I have simply corrected syntax errors in your program, but I don't think it will actually get you E. See this page about calculating E in C.

Answer 2

I'm no C master but isnt just calling double by itself a type declaration and not type casting? wouldnt it be power = (double) pow(x, x); if you are type casting? see: https://www.tutorialspoint.com/cprogramming/c_type_casting.htm

Answer 3

I reworked some mistakes in your code and think it should work now; however, the style, which I kept untouched, is confusing.
#include <stdio.h>
#include <math.h>
int main()
{
double power; //stores floating point numbers!
double num = 1;//stores floating point numbers!
int position = 1;
while (position <= 100)
{
num = 1/num;
num = num + 1;
power = pow(num, position); //x not needed, this is what you ment
printf("%f\n", power); //%d outputs decimal numbers, f is for floats
position += 1;
num = position;
}
}
To improve your code, I would suggest to simplify it. Something along the lines of this
#include <stdio.h>
#include <math.h>
int main()
{
double approx;
for(int iter=1; iter<=100; iter++){
approx=pow((1+1./iter),iter);
printf("%f\n", approx);
}
}
is much easier to understand.