2 ms·
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?
by tgv 5y 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 5y agoNo, not in Fortran of that era ~ was pretty much missing from most card punch keyboards
- deleted 5y ago[deleted]