C++ Floating Point Arithmetic Floating Point Numbers are Weird
Example
The first mistake that nearly every single programmer makes is presuming that this code will work as intended:
float total = 0;
for(float a = 0; a != 2; a += 0.01f) {
total += a;
}
The novice programmer assumes that this will sum up every single number in the range 0, 0.01, 0.02, 0.03, ..., 1.97, 1.98, 1.99, to yield the result 199—the mathematically correct answer.
Two things happen that make this untrue:
- The program as written never concludes.
anever becomes equal to2, and the loop never terminates. - If we rewrite the loop logic to check
a < 2instead, the loop terminates, but the total ends up being something different from199. On IEEE754-compliant machines, it will often sum up to about201instead.
The reason that this happens is that Floating Point Numbers represent Approximations of their assigned values.
The classical example is the following computation:
double a = 0.1;
double b = 0.2;
double c = 0.3;
if(a + b == c)
//This never prints on IEEE754-compliant machines
std::cout << "This Computer is Magic!" << std::endl;
else
std::cout << "This Computer is pretty normal, all things considered." << std::endl;
Though what we the programmer see is three numbers written in base10, what the compiler (and the underlying hardware) see are binary numbers. Because 0.1, 0.2, and 0.3 require perfect division by 10—which is quite easy in a base-10 system, but impossible in a base-2 system—these numbers have to be stored in imprecise formats, similar to how the number 1/3 has to be stored in the imprecise form 0.333333333333333... in base-10.
//64-bit floats have 53 digits of precision, including the whole-number-part.
double a = 0011111110111001100110011001100110011001100110011001100110011010; //imperfect representation of 0.1
double b = 0011111111001001100110011001100110011001100110011001100110011010; //imperfect representation of 0.2
double c = 0011111111010011001100110011001100110011001100110011001100110011; //imperfect representation of 0.3
double a + b = 0011111111010011001100110011001100110011001100110011001100110100; //Note that this is not quite equal to the "canonical" 0.3!
