What is the correct way to obtain (-1)^n?

Question

Many algorithms require to compute (-1)^n (both integer), usually as a factor in a series. That is, a factor that is -1 for odd n and 1 fo

Answer 1

First of all, the fastest isOdd test I do know (in an inline method)

/**
* Return true if the value is odd
* @value the value to check
*/
inline bool isOdd(int value)
{
return (value & 1);
}

Then make use of this test to return -1 if odd, 1 otherwise (which is the actual output of (-1)^N )

/**
* Return the computation of (-1)^N
* @n the N factor
*/
inline int minusOneToN(int n)
{
return isOdd(n)?-1:1;
}

Last as suggested @Guvante, you can spare a multiplication just flipping the sign of a value (avoiding using the minusOneToN function)

/**
* Example of the general usage. Avoids a useless multiplication
* @value The value to flip if it is odd
*/
inline int flipSignIfOdd(int value)
{
return isOdd(value)?-value:value;
}