Example 3 is not correct. The modulo operator is not branchless. The only way to make this branchless is to have the second argument be a power of 2 (in this case, modulos can be calculated by bit-shifting). Instead, do something like:
return x[i - (i >= lenX) * lenX]
Of course, we are assuming i is less than lenX
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Example 3 is not correct. The modulo operator is not branchless. The only way to make this branchless is to have the second argument be a power of 2 (in this case, modulos can be calculated by bit-shifting). Instead, do something like:
return x[i - (i >= lenX) * lenX]Of course, we are assuming i is less than lenX