Comparing doubles

Question

I'm writing a program that consists of a while loop that reads two doubles and prints them. The program also prints what the larger number is and what the smaller number is.

this is the code i have so far.

int main()
{

                                    // VARIABLE DECLARATIONS 

    double a;
    double b;

    while (a,b != '|')              //WHILE A & B DO NOT EQUAL '|'
    {
        cin >>a >>b;
        cout << a << b << "\n" ;


        if (a<b)                    //IF A<B: SMALLER VALUE IS A
        cout << "The smaller value is:" << a << endl 
             << "The larger value is:" << b << endl ;

        else if (b<a)               //ELSE IF B<A 
            cout << "The smaller value is:" << b << endl 
                 << "The larger value is:" << a << endl ;
        else if (b==a)
            cout << "The two numbers you entered are equal." << "\n" ;

    }
}

The next step is having the program write out "the numbers are almost equal" if the two numbers differ by less than 1.0/10000000. How would I do this?

Answer 1

std::abs(a - b) < 0.000001

Of course, replace the constant with whatever you consider "almost".

Answer 2

Just test if they differ by less than that amount :)

if ( std::abs(a - b) < 1.0 / 10000000 )
  cout << "The numbers are almost equal.\n";

Answer 3

if (a * 1.0000001 > b  &&  a < b*1.0000001)

You can add an error value (your 1.0 / 10000000.0 ) but it is generally better to use a multiplier, so the comparison is to the same level of accuracy.

Answer 4

abs(a - b) < 1.0 / 10000000

Answer 5

Here is how I would test for equality, without a large "fudge factor":

if (std::abs(a-b)<2.0*std::numeric_limits<double>::epsilon())
  std::cout << "The numbers are equal\n";

Or, if a and b are relatively large, you can expand this, to account for the (floor of the) maximum digits (≈15.955) that a double can represent:

if (std::abs(a-b)<2.*std::numeric_limits<double>::epsilon() || // For small a, b
    std::abs(a-b)<std::min(a,b)/1.e-15) // For large a, b
  std::cout << "The numbers are equal\n";

Finally, I recommend that you rethink your while (a,b != '|') construct, because it most likely does not do what you think.

Answer 6

If you want the test to scale with a and b, you could try testing abs(a/b-1) < e, where e is your favorite tiny number, like 0.001. But this condition is actually asymmetrical in a and b, so it can work out to say a is close to b, but b is not close to a. That would be bad. It's better to do abs(log(a/b)) < e, where e, again, is your favorite tiny number. But the logarithms present extra computation, not to mention terrifying undergraduates everywhere.

Answer 7

I suggest the following article: Comparing floating point numbers