Unexpected output using the switch statement for this complex logic

I am trying to use a switch s开发者_开发知识库tatement for 3 conditions. Conditions were:

1. When a, b, and c are all zero, any value of x is a solution. Print: Any value of x is a solution.
2. When a and b are zero and c is not, no solution exists. Print: No solution exists.
3. When a is zero and b is not zero, the only solution is x = -c/b. Calculate the value of x and print the solution.

When I tried to run my program, it displayed the wrong results. My input were

a = 0
b = 0
c = 0

So it's supposed to print "Any value of x is a solution", but it didn't.

My program is:

#include <stdio.h>

//Function declarations
void getData (int* a, int* b, int* c);
float calculateX (int a, int b, int c);

//===================================================
int main (void)
{
    //Local declarations
    int a;
    int b;
    int c;
    float x;

    //Statements
    getData (&a, &b, &c);
    calculateX (a, b, c);

    int temp;
    printf("\nEnter an integer and press enter to exit the program: ");
    scanf("%d", &temp);

    return 0;
}

//----------------------------------------------------
void getData (int* a, int* b, int* c)
{
    printf("Enter three integers: ");
    scanf("%d %d %d", a, b, c);
    return;
}

//----------------------------------------------------
float calculateX (int a, int b, int c)
{
    float x;

    printf("Input is: %d %d %d\n", a, b, c);
    switch(a, b, c)
    {
        case 1: (a, b, c == 0);
                printf("Any value of x is a solution.");
                break;
        case 2: (a, b == 0 && c!= 0);
                printf("No solution exists.");
                break;
        case 3: (a == 0 && b!= 0);
                x = (float)(-c/b);
                printf("The value of x is: %.1f", x);
                break;
        default: printf("Cannot calculate.");
    }
    return a, b, c;
}

And my output was:

Enter three integers: 0 0 0
Input is: 0 0 0
Cannot calculate.
Enter an integer and press enter to exit the program:

This is not how a switch statement works. It compiles, but for very obscure reasons. Obviously, it doesn't do what you expect when it runs.

Generally speaking, you use a switch statement on a single expression, and each of the case labels represents one possible value of that expression. e.g.:

switch (x)
{
case 1:
    // Code here runs when x == 1
    break;
case 2:
    // Code here runs when x == 2
    break;
default:
    // Code here runs for all other values of x
    break;
}

In your application, you want to test multiple variables, and combine them in complex ways. There is no neat way to do that with switch. You should consider a set of if statements instead.

Is there any reason you have to use switch? Just do

if (a == 0 && b == 0 && c == 0)
  ...
else if (a == 0 && b == 0 && c != 0)
  ....
...

Actually, here's a valid way to use the switch statement for this problem:

switch ((a != 0) * 4 + (b != 0) * 2 + (c != 0))
{
case 0: // a, b, c == 0
    printf("Any value of x is a solution.");
    break;

case 1: // a, b == 0 && c!= 0
    printf("No solution exists.");
    break;

case 2: // a == 0 && b!= 0
case 3:
    x = (float)(-c/b);
    printf("The value of x is: %.1f", x);
    break;

default:
    printf("Cannot calculate.");
}

I've based this on your code, except that I use conditionals (which evaluate to 0 or 1) to encode the state of each variable (zero or not, respectively) in the expression, assigning each to a separate bit. The switch then decodes it -- the interesting part for you, as a beginner, is that case 2 falls through to case 3 because we don't care whether c is zero.

Your code has some other issues, but I'm restricting myself to the switch that you asked about. Best of luck.