4 ms·

does it work for unary +/-?

• truly 4y ago
  Almost: +13 would be rewritten to ((()))+(((13))) and -13 to ((()))-(((13))). If you interpret the empty expression as 0, perhaps it would work out. Good question!
  • tgv 4y ago
    There are other unary operators (!, ~), although I'm almost sure FORTRAN didn't support those, but 1 - -13 => 1 - 0 - 13 = -12 would fail, wouldn't it?
    • Taniwha 4y ago
      No, not in Fortran of that era ~ was pretty much missing from most card punch keyboards
    • deleted 4y ago
      [deleted]