Floating-point traps

1#include <stdio.h>2 intMain ()3 {4     Doublei;5      for(i =0.0; I! =Ten; i + =0.1)6printf"%lf\n", i);7     return 0;8}

Will this program stop at 10??? The answer is no.

Causes of the above traps

In fact, all data in the computer is represented by a binary representation, including floating-point numbers. This causes some floating-point numbers not to be represented in binary precision, such as 0.1 (which is easy to understand, as 10/3 cannot be represented by a decimal)

Further, floating-point numbers are represented by a fraction + exponent, for example

0.5 = 1/2

0.75 = 1/2 + 1/(2^2)

0.875 = 1/2 + 1/(2^2) + 1/(2^3)

0.1 = 1/(2^4) + 1/(2^5) + 1/(2^8) + ...

0.1 of them can loop indefinitely, which means that 0.1 cannot be accurately represented in a computer, so creating these traps is easy to understand.

Traps for floating-point numbers