Floating point Trap

Trap One

if (0.1+0.1+0.10.3)      printf ("equal\n" );   Else       printf ("notequal\n");  

Does it output "equal"? No! Actual floating-point arithmetic,0.1+0.1+0.1! = 0.3

Trap Two

Double i;  for (i=0; i!=; i+=0.1)    printf ("%.1lf\n", i);

Will the program end the loop from 0.0 to 9.9? No! Because I can never be equal to ten, this will be infinite loop

Causes of the above traps

All data in the computer is represented in binary notation, including floating-point numbers, which causes some floating-point numbers not to be represented by binary precision, such as 0.1(which is easy to understand, like 10/3=0.33333 ...). cannot be represented exactly as a decimal)

For example:

(0.5) 10 = 2-1 = (0.1) 2

(0.75) 10 = 2-1 + 2-2 = (0.11) 2

(0.875) 10 = 2-1 + 2-2 + 2-3 = (0.111) 2

(0.1) 10 = 2-4 + 2-5 + 2-8 + ... = (0.00011001 ...) 2

0.1 of them can loop indefinitely, which means that 0.1 cannot be accurately represented in the computer, so it is easy to understand that these two traps are involved in a 0.1 participation operation that cannot be represented by a binary precision, so the expected results are not obtained. For example, trap two, change the step size to a binary precise representation of 0.5, will not be infinite loop down.

Floating point Trap