My first program in C. A simple command-line calculator

Question

I've recently started programming in C and as my first serious program I thought I'd create a simple calculator. To make it a bit more complex I decided not to use the functions included in math.h (like pow) but to create these functions by myself. Let me know if there is something I can do to improve this code.

#include<stdio.h>


double power(double x, double y, double r)                                                       // This function calculates the power and takes numer1, number2 and the results
{
    r = 1;                                                                           // We need to initialize the results otherwise we can't store the partial result on line 11
    int c;  
    for (c = y; c > 0; c--)                                  // putting c equal to y we can decrement c and obtain the precise number of operations needed to complete the power
    { 
        int pow = y;                                  // that's a bit tricky: we can't apply the modulus operand in line 12 because y is a double, so we must convert y in an int 
        r *= x;                                                             // this way result = 1 * number1, then number1 * number1, then number1 * number1 * number1 and so on
        if ((x < 0) && (pow % 2 == 1))    /* If the first number is negative, and we calculate the power of this number to 2 or multiples, the number resulting will be negative
                                                                to obtain the correct result (positive) we must multiply the result to -1, if the previous conditions are met */
        {
            x *= -1;
        }
    }
    printf("%0.1lf", r);
}


double root(double x, double y, double r)
{
    int c;
      if (x < 0)
      {
         puts("is impossible to calcultate the root of a negative number");
      }
    for (c = 1; c < x; c++)
    {
        int partRes = x;
        partRes /= c;
        if (partRes * partRes == x)
        {
            printf("The root of %0.1lf is: %0.1d\n", x, partRes);
            break;
        }
    }
}


int main()
{
    double num1, num2, res;
    char op;
    puts("Input number, operator and number");
    puts("operands permitted: + (addition), - (subtraction), * (multiplication), / (division), ^ (power), $ (square root)");
    scanf("%lf %c %lf", &num1, &op, &num2);
    if (op == '+')
    {
        res = num1 + num2;
        printf("%0.1lf", res);
    }
    else if (op == '-')
    {
        res = num1 - num2;
        printf("%0.1lf", res);
    }
    else if (op == '*')
    {
        res = num1 * num2;
        printf("%0.1lf", res);
    }
    else if (op == '/')
    {
        res = num1 / num2;
        printf("%0.1lf", res);
    }
    else if (op == '^')
    {
        power(num1, num2, res);
    }
    else if (op == '$')
    {
        root(num1, num2, res);
    }
    else 
    printf("error, use a valid operand");
}

Answer 1

power() problems

double power() does not return a value.

Incorrect result when y is not a value in the int range.

Code attempts a linear power calculation: O(y). Research Exponentiation by squaring. It is faster and less precision loss.

double r parameter serves no purpose. Instead make a local double r variable.

A floating point power function is one of the most changeling functions to code well. Rough cut below.

---

root() problems

double root() does not return a value.

Other problems like power().

---

It floating point, not fixed point

More informative to use exponential notation for large and small value than a fix-point format.

// printf("%0.1lf", res);
printf("%g", res);

---

Sample improved my_power() - unchecked

// Very rough code
double my_power(double x, double y) {
  double whole;
  double fraction = modf(y, &whole);
  if (fraction == 0.0) {
    double r = 1.0;
    double p = fabs(whole);
    while (p > 0) {
      if (fmod(p, 2.0)) {
        r *= x;
      }
      x *= x;
      p /= 2.0;
    }
    if (whole < 0) {
      r = 1.0 / r;
    }
    return r;
  }
  if (x == 0.0) {
    return 0.0;
  }
  if (x < 0.0) {
    return NAN;
  }
  return exp(log(x) * y);
}