Using "double" as counter variables in loops

Question

In a book I am currently reading, there is this excerpt:

You can also use a floating-point value as a loop counter. Here's an example of a for loop with this kind of counter:

double a(0.3), b(2.5);
for(double x = 0.0; x <= 2.0; x += 0.25)
    cout << "\n\tx = " << x << "\ta*x + b = " << a*x + b;

This code fragment calculates the value of a*x+b for values of x from 0.0 to 2.0, in steps of 0.25; however, you need to take care when using a floating-point counter in a loop. Many decimal values cannot be represented exactly in binary floating-point form, so discrepancies can build up with cumulative values. This means that you should not code a for loop such that ending the loop depends on a floating-point loop counter reaching a precise value. For example, the following poorly-designed loop never ends:

for(double x = 0.0 ; x != 1.0 ; x += 0.2)
    cout << x;

The intention with this loop is to output the value of x as it varies from 0.0 to 1.0; however, 0.2 has no exact representation as a binary floating-point value, so the value of x is never exactly 1. Thus, the second loop control expression is always false, and the loop continues indefinitely.

Can someone please explain how the first code block runs while the second doesn't?

Answer 1

The first one will eventually terminate, even if x doesn't reach exactly 2.0... because it'll end up being greater than 2.0, and thus break out.

The second one would have to make x hit exactly 1.0 in order to break.

It's unfortunate that the first example uses a step of 0.25, which is exactly representable in binary floating point - it would have been smarter to make both examples use 0.2 as the step size. (0.2 isn't exactly representable in binary floating point.)