Faster way to do the saturation(n) function?

Asked Jun 01 '15 at 16:23

Active Jun 01 '15 at 17:05

Viewed 42 times

I want to do the saturation function as following:

/* Ensure number is >=0 and <=255 */
#define Saturate(n) ((n) > 0 ? ((n) < 255 ? (n) : 255) : 0)

If 0 < n < 255, return n; if n <= 0, return 0; if n >= 255, return 255.

I am looking for faster way of doing it. For example, is there any built-in function in c which can be used to reduce the number of instructions?

edited Jun 01 '15 at 17:05

sepp2k

asked Jun 01 '15 at 16:23

fluency03

Take a look at Trap's answer on [this question](http://stackoverflow.com/questions/3176602/how-to-force-a-number-to-be-in-a-range-in-c) – John Odom Jun 01 '15 at 16:30
For gcc, there are builtins for certain architectures. However, they are not portable. Which type is n? You reall yshould use functions, not defines. `inline` functions are as good as macros here, but conserve type information and a re easier to debug. – too honest for this site Jun 01 '15 at 16:37
(n>255)*255 + (1-(n<0 || n>255))*n – huseyin tugrul buyukisik Jun 01 '15 at 20:53
`if (a&(~0xFF)) return (-a)>>31; else return a;` – fluency03 Jun 02 '15 at 01:04

0 Answers0