How to wrap a number using the mod operator

Question

Not sure if this is possible, but is there an automatic way using mod or something similar to automatically fix bad input values? For instance:

If r>255, then set r=255 and
if r<0, then set r=0

So basically what I ask is a smart math way to fix this and not use

if(r>255)
 r=255;
if(r<0)
 r=0;

Answer 1

What about:

r = std:max(0, std::min(r, 255));

Answer 2

The following function displays what you are looking for:

 f(x) = (510*(1 + Sign[-255 + x]) + x*(1 + Sign[255 - x])*(1 + Sign[x]))/4

As below:

Answer 3

- :

R = MIN(r, 255);
R = MAX(R, 0);

Answer 4

, , , ints, :

, unsigned int - 16 ( ):

r = r & 0000000011111111;

int 32 , 16 -.

r 255. int - . , ( , , ). , r = min(r, 0); .

Answer 5

. ( 0 255) :

static inline int trim_8bit(unsigned i){
    return 0xff & ((i | -!!(i & ~0xff))) + (i >> 31);
    // where "0xff &" can be omitted if you return unsigned char
};

,

static inline unsigned char trim_8bit_v2(unsigned i){   
    if (__builtin_expect(i & ~0xFF, 0)) // it for gcc, use __assume for MSVC
        return (i >> 31) - 1; 
    return i;
};

, , .