My for-loop does not start [duplicate] - c++

Question

This question already has answers here:
Compare double to zero using epsilon
(12 answers)
Closed 8 years ago.
I know there are loads of topics about this question, but none of those helped me. I am trying to find the root of a function by testing every number in a range of -10 to 10 with two decimal places. I know it maybe isn't the best way, but I am a beginner and just want to try this out. Somehow the loop does not work, as I am always getting -10 as an output.
Anyway, that is my code:
#include <iostream>
using namespace std;
double calc (double m,double n)
{
double x;
for (x=-10;x<10 && m*x+n==0; x+=0.01)
{
cout << x << endl;
}
return x;
}
int main()
{
double m, n, x;
cout << "......\n";
cin >> m; // gradient
cout << "........\n";
cin >> n; // y-intercept
x=calc(m,n); // using function to calculate
cout << ".......... " << x<< endl; //output solution
cout << "..............\n"; // Nothing of importance
return 0;
}

Answer 1

You are testing the conjunction of two conditions in your loop condition.
for (x=-10;x<10 && m*x+n==0; x+=0.01
For many inputs, the second condition will not be true, so the loop will terminate before the first iteration, causing a return value of -10.
What you want is probably closer to something closer to the following. We need to test whether the absolute value is smaller than some EPSILON for two reasons. One, double is not precise. Two, you are doing an approximate solution anyways, so you would not expect an exact answer unless you happened to get lucky.
#define EPSILON 1E-2
double calc (double m,double n)
{
double x;
for (x=-10;x<10; x+=0.001)
{
if (abs(m*x+n) < EPSILON) return x;
}
// return a value outside the range to indicate that we failed to find a
// solution within range.
return -20;
}
Update: At the request of the OP, I will be more specific about what problem EPSILON solves.
double is not precise. In a computer, floating point number are usually represented by a fixed number of bits, with the bit representation usually being specified by a standard such as IEE 754. Because the number of bits is fixed and finite, you cannot represent arbitrary precision numbers. Let us consider an example in base 10 for ease of understanding, although you should understand that computers experience a similar problem in base 2.
If m = 1/3, x = 3, and n = -1, we would expect that m*x + n == 0. However, because 1/3 is the repeated decimal 0.33333... and we can only represent a fixed number of them, the result of 3*0.33333 is actually 0.999999, which is not equal to 1. Therefore, m*x + n != 0, and our check will fail. Thus, instead of checking for equality with zero, we must check whether the result is sufficiently close to zero, by comparing its absolute value with a small number we call EPSILON. As one of the comments pointed out the correct value of EPSILON for this particular purpose is std::numeric_limits::epsilon, but the second issue requires a larger EPSILON.
You are are only doing an approximate solution anyways. Since you are checking the values of x at finitely small increments, there is a strong possibility that you will simply step over the root without ever landing on it exactly. Consider the equation 10000x + 1 = 0. The correct solution is -0.0001, but if you are taking steps of 0.001, you will never actually try the value x = -0.0001, so you could not possibly find the correct solution. For linear functions, we would expect that values of x close to -0.0001, such as x = 0, will get us reasonably close to the correct solution, so we use EPSILON as a fudge factor to work around the lack of precision in our method.

Answer 2

m*x+n==0 condition returns false, thus the loop doesn't start.
You should change it to m*x+n!=0