Need explanation: Equation solving code

[Nielyboyken]

Hello

A few months ago I found a code which was able to solve equations.
At first I didn't understood it at all, I tried examining the code and with some support I now understand how the operators, structs, overloads, hooking (with line 59 - via overloading),... works

I accomplished to understand the whole code, EXCEPT the mathematical and logical part - the actually most important part -.

I'd really appreciate it, if someone was able to explain me what the math-part in the constructors +,-,/,* do, and work.

How is this piece code for example, able to sum up two terms?

LINE operator + (LINE A,LINE B) {	//assumes that A.x == 0 or B.x == 0 or A.x == B.x
	LINE R;
	R.a = A.a + B.a;
	R.k = A.k + B.k;
	if(A.x) R.x = A.x;
	else R.x = B.x;
	return R;
}

Full code:

#include <iostream>
#include <math.h>

struct VAR{
	float i;
};

struct LINE{	//k*x+a
	float a,k;
	VAR* x;

	LINE(){}
	LINE(int a): a(a),k(0),x(0) {}
	LINE(VAR& v): a(0),k(1),x(&v) {}
};

LINE operator + (LINE A,LINE B) {	//assumes that A.x == 0 or B.x == 0 or A.x == B.x
	LINE R;
	R.a = A.a + B.a;
	R.k = A.k + B.k;
	if(A.x) R.x = A.x;
	else R.x = B.x;
	return R;
}

LINE operator - (LINE A,LINE B) {	//same as +
	LINE R;
	R.a = A.a - B.a;
	R.k = A.k - B.k;
	if(A.x) R.x = A.x;
	else R.x = B.x;
	return R;
}

LINE operator * (LINE A,LINE B) {	//assumes that A.x == 0 or B.x == 0
	LINE R;
	R.a = A.a * B.a;
	R.k = A.k * B.a + B.k * A.a;
	if(A.x) R.x = A.x;
	else R.x = B.x;
	return R;
}

LINE operator / (LINE A,LINE B) {	//assumes that B.x == 0
	LINE R;
	R.a = A.a / B.a;
	R.k = A.k / B.a;
	R.x = A.x;
	return R;
}

void operator == (LINE A,LINE B)	{
	LINE C = A - B;
	C.x->i = -C.a/C.k;
}

int main(){
	VAR x;
	4*x == 2*2;

	std::cout << "x = " << x.i;
	std::cin.get();

	return 0;
}

Other things which would be nice to have it explained:
- Line 12 - 14 (the last three lines of where the struct LINE is initialised),
- Some more info about the operator overloading part here,
- More explanation about the constructors used in this code;

I'd really like to understand how this code works.
Thanks a lot all!

[keanedawg]

What have you found?

It looks like some sort of mx + b type of equation. Though I'm not really sure beyond that. What exactly is the program this struct is used in?

[YFGHNG]

LINE operator + (LINE A,LINE B) {	//assumes that A.x == 0 or B.x == 0 or A.x == B.x
	LINE R;
	R.a = A.a + B.a;
	R.k = A.k + B.k;
	if(A.x) R.x = A.x;
	else R.x = B.x;
	return R;
}

Line 1: operator+ first takes in two parameters of object LINE

Line 2: declare a local LINE object

Line 3: call the constructor and initialize values to float a

Line 4: same thing as line 3, but with k and 0

Line 5-6: I'm thinking that's for if A.x has a value other than 0, then do first line, otherwise do the else

- Line 12 - 14 (the last three lines of where the struct LINE is initialised),
- Some more info about the operator overloading part here,
- More explanation about the constructors used in this code;

Last 3 lines are just different constructors. The colon and all the variables with the parentheses are just shorthand for initializing variables to a certain value within the constructor.

Operators are functions too, if you don't define them for custom object classes, you can't use them. You can't call int multiply( int n ) and then try to insert a string in there.

[Nielyboyken]

@Kean: the full code is posted above.
The equation in this code:
4x = 2*2

The result is of course 1.

@YFGHN:
Thanks a lot for your explanation!
Still I'd like to know a step-by-step explanation on how it's calculated. How does this program calculates the equation from the equation to finding 'x'.

[TwilightSpectre]

Sometimes it can help to just step through the code and work out what's going on.

On line 59, 4 and x get implicitly converted to LINEs so as to provide a valid multiplication/equality operator, and the same for the 2*2 on the other side.

So, on one side we have a LINE with the value a=0;k=4, and on the other side wie have a LINE with the value a=4;k=0. That is, we have represented the equation 4x=4.

Now, to work out x in this circumstance is down to the == operator. We get the LINE C, with values a=-4;k=4. Finally, we set the value of our variable x to -(-4)/4, and we get the result 1.

---

Basically, each LINE has a variable which is passed around as a pointer, and whenever we do an operation on it we modify k (the coefficient) and a (the offset).

Finally, at the end it all culminates with k1*x+a1 == k2*x+a2, where we move all x to one side and all a to the other, and then store the result in our pointer to x, which we can retrieve at our leisure.

Does that help?