How to optimise my C code to solve quadratic equations in R?

Question

I just wanna know how I can optimize my C code. My program works fine, I tested it with many different values, all is good. However, I'd like to reduce the number of lines and write my program in better quality. Here's the source code:

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

int main(void) {
    float a,b,c,x,x1,x2;
    printf("aX^2 + bX + c = 0\n");
    printf("Type the value of a: ");
    scanf("%f", &a);
    printf("Type the value of b: ");
    scanf("%f", &b);
    printf("Type the value of c: ");
    scanf("%f", &c);

    if ( a!=0 && b!=0 && c!=0){
        float delta = b*b - 4*a*c;
        if (delta>0){
            x1 = (-b-sqrt(delta))/(2*a);
            x2 = (-b+sqrt(delta))/(2*a);
            printf("Solutions are x1 = %f and x2 = %f\n",x1,x2);
        }
        else if (delta == 0){
            x = -b/(2*a);
            printf("One unique solution is x = %f\n", x);
        }
        else {
            printf("No solutions !\n");
        }
    }
    if ( a==0 && b!=0 && c!=0)
        printf("One unique solution x = %f\n", -c/b);
    if ( a==0 && b==0 && c!=0)
        printf("No solutions !\n");
    if ( a==0 && b==0 && c==0 )
        printf("Set of solutions is R\n");
    if (a!=0 && b==0 & c!=0) {
        x = -c/a;
        if(x>=0)
            printf("Two soltions x = %f et -x = %f\n", sqrt(x),-sqrt(x));
        else{
            printf("No solutions !\n");
        }
    }
    if (a!=0 && b==0 && c==0)
        printf("One unique x = 0\n");

}

Answer 1

One small optimization: use else if instead of all raw ifs. This will avoid unnecessary comparisons after one if condition is matched.

You can also experiment with the order of the if statements. If any conditions are more often to be true for your use cases, then they should be first.

Answer 2

Since you are basically building a state machine, let's actually do so:

int state = (a == 0) ? 0 : 1;
state |= (b == 0) ? 0 : 2;
state |= (c == 0) ? 0 : 4;

switch(state)
{
    case 0: // All co-efficents 0 Domain R print 
        break; 
    case 1: // Only a coefficient: 0 case
        break;
    case 2: // Only b coefficient 0 case.
        break;
    case 3: // a and b coefficient case - one root 0, other linear.
        break;
    case 4: // Only c case - no solutions
        break;
    case 5: // a and c case
        break;
    case 6:  // b and c case (linear)
        break;
    case 7: // a,b,c all here - do full quadratic solve.
         break;
    default:
       // This should never happen - assert and exit.
         break;
}

This way, all your cases are explicitly defined. Each case could also then call a function. Or, one step further, an array of function pointers for each case, fully and properly named.

Answer 3

1) Some of the parts of calculations are being repeated. Break the calculations into smaller pieces, store the results in variables, and use them to build your needed results.

2) You can avoid the special handling of the case a!=0 && b==0 & c!=0 entirely (albeit you need to change some of the other tests).

3) Some operations (e.g. prompting and reading) are repeated. Do these in a function which takes an argument (e.g. a string for prompt).

4) The first line where you are printing "No solutions" there are actually complex or imaginary solutions. The second place is another type of solution.

5) Possibly break some of the repeated calculations (point 1) into separate functions too.

6) Slightly advanced: your code will not deal well with a user who enters rubbish (non-numeric) input.

Answer 4

You can use this example to optimise your code:

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

int main()
{
    double a, b, c, determinant, root1,root2, realPart, imaginaryPart;

    printf("Enter coefficients a, b and c: ");
    scanf("%lf %lf %lf",&a, &b, &c);

    if(a==0 && b!=0 && c!=0){
     root1 = -c/b;
     printf("linear equation root = -c/b = %.2lf", root1);
    }else{

    determinant = b*b-4*a*c;

    // condition for real and different roots
    if (determinant > 0)
    {
    // sqrt() function returns square root
        root1 = (-b+sqrt(determinant))/(2*a);
        root2 = (-b-sqrt(determinant))/(2*a);

        printf("root1 = %.2lf and root2 = %.2lf",root1 , root2);
    }

    //condition for real and equal roots
    else if (determinant == 0)
    {
        root1 = root2 = -b/(2*a);

        printf("root1 = root2 = %.2lf;", root1);
    }

    // if roots are not real 
    else
    {
        realPart = -b/(2*a);
        imaginaryPart = sqrt(-determinant)/(2*a);
        printf("root1 = %.2lf+%.2lfi and root2 = %.2f-%.2fi", realPart, imaginaryPart, realPart, imaginaryPart);
    }
}
    return 0;
}

• If determinant is greater than 0, the roots are real and different.
• If determinant is equal to 0, the roots are real and equal.
• If determinant is less than 0, the roots are complex and different.

So,

Sample image