2 ms·

Mathematica has symbolic and infinite-precision addition, so you can't automatically take advantage of obvious compiled code.

• sfpotter 7mo ago
  What? Arbitrary precision arithmetic implemented in a compiled language will be faster than the alternative. This is no great mystery. The same is true of essentially all low-level symbolic or numerical math algorithms. You need to get to a fairly high level before this stops being true.
  • creato 7mo ago
    Of course. The point is whether you interpret a call to arbitrary_precision_add or compile the call doesn't matter much.
    • deleted 7mo ago
      [deleted]