
Re: Reason for operator precedence
Posted:
Mar 15, 2006 8:56 AM


In article <dv6nv8$chc$1@agate.berkeley.edu> magidin@math.berkeley.edu (Arturo Magidin) writes: ... [ Without operator precedence ] ... > If you want to put in the square, you > would need to do something like > > bx + c + (a(x^2)). > > Higher degree polynomials would be even harder.
It can be done: a * x + b * x + c (thanks to Horner), but it is not immediately clear that it is a polynomial.  dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131 home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/

