get absolute value without using abs function nor if statement

Branchless:

int abs (int n) {
    const int ret[2] = { n, -n };
    return ret [n<0];
}

_{Note 4.7 Integral Conversions / 4: [...] If the source type is bool, the value false is converted to zero and the value true is converted to one.}

I try this code in C, and it works.

int abs(int n){
   return n*((2*n+1)%2); 
}

Hope this answer will be helpful.

From Bit Twiddling Hacks:

int v;           // we want to find the absolute value of v
unsigned int r;  // the result goes here 
int const mask = v >> sizeof(int) * CHAR_BIT - 1;

r = (v + mask) ^ mask;

int abs(int v) 
{
  return v * ((v>0) - (v<0));
}

This code multiplies the value of v with -1 or 1 to get abs(v). Hence, inside the parenthesis will be one of -1 or 1.

If v is positive, the expression (v>0) is true and will have the value 1 while (v<0) is false (with a value 0 for false). Hence, when v is positive ((v>0) - (v<0)) = (1-0) = 1. And the whole expression is: v * (1) == v.

If v is negative, the expression (v>0) is false and will have the value 0 while (v<0) is true (value 1). Thus, for negative v, ((v>0) - (v<0)) = (0-1) = -1. And the whole expression is: v * (-1) == -v.

When v == 0, both (v<0) and (v>0) will evaluate to 0, leaving: v * 0 == 0.