Running Quadratic Equation solver in C, on Linux - c

Question

I'm a little new to programming, I'm trying to write a program to solve quadratic equation, here's my code:
#include <stdio.h>
#include <math.h>
void main (){
int a, b, c, delta;
float root1, root2;
delta = ( b * b ) - ( 4 * a * c );
printf("enter a, b, c, in such syntax ax^2 + bx + c:\n");
scanf("%d%d%d", &a, &b, &c);
printf("You mean %dx^2 + %dx + %d, delta=%f\n\n", a, b, c, delta);
if ( delta < 0 )
printf("The equation has no roots.\n");
if ( delta == 0 ){
root1 = -b / (2*a);
printf("The equaion has one root: %d\n", root1);
}
if ( delta > 0 ){
root1 = (-b + sqrt(delta)) / (2*a);
root2 = (-b - sqrt(delta)) / (2*a);
printf("root 1 = %f\nroot 2 = %f\n", root1, root2);
}
}
It's compiled without error, the problem is each time I run it, with same input, I get different answers!
$ ./qe
enter a, b, c, in such syntax ax^2 + bx + c:
2
4
2
You mean 2x^2 + 4x + 8, delta=0.000000
The equation has no roots.
$ ./qe
enter a, b, c, in such syntax ax^2 + bx + c:
2
4
2
You mean 2x^2 + 4x + 8, delta=0.000000
root 1 = 6543.122070
root 2 = -6545.122070
$ ./qe
enter a, b, c, in such syntax ax^2 + bx + c:
2
4
2
You mean 2x^2 + 4x + 8, delta=0.000000
root 1 = 8342.037109
root 2 = -8344.037109
$ ./qe
enter a, b, c, in such syntax ax^2 + bx + c:
2
4
2
You mean 2x^2 + 4x + 8, delta=0.000000
The equation has no roots.
.
.
I'm so confused. Whats the problem? My gcc version: 6.3.1 20170306 (GCC)
how should I fix this?
Thanx.

Answer 1

TL;DR Your code causes undefined behavior.
To elaborate, in the case
delta = ( b * b ) - ( 4 * a * c );
a, b, c are used uninitialized. They
can have trap representation
do not have their address taken.
So, this invoke UB.
You need to
first check for the success of scanf() call
move the calculation of delta after the successful scanning for values for the involved operands.
Also, to add, for a hosted environment, the conforming signature for main() would be int main(void), not void main ()