# Floating point arithmetic

suggest changeFloating point numbers are weird.

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.
`a`

never becomes equal to`2`

, and the loop never terminates. - If we rewrite the loop logic to check
`a < 2`

instead, the loop terminates, but the total ends up being something different from`199`

. On IEEE754-compliant machines, it will often sum up to about`201`

instead.

The reason that this happens is that **Floating Point Numbers represent approximations of their assigned values**.

The classical example is the following computation:

```
#include <iostream>
#include <string>
int main(int argc, char **argv) {
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;
}
}
```

```
This Computer is pretty normal, all things considered.
```

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!
```