11 Nov 2006 21:30:49 -0500

Related articles |
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Fastening the run-time interpretation of mathematical expressions paolopantaleo@gmail.com (PAolo) (2006-11-11) |

Re: Fastening the run-time interpretation of mathematical expressions haberg@math.su.se (2006-11-11) |

Re: Fastening the run-time interpretation of mathematical expressions mailbox@dmitry-kazakov.de (Dmitry A. Kazakov) (2006-11-13) |

Re: Fastening the run-time interpretation of mathematical expressions akk@privat.de (Andreas Kochenburger) (2006-11-13) |

Re: Fastening the run-time interpretation of mathematical expressions martin@gkc.org.uk (Martin Ward) (2006-11-13) |

Re: Fastening the run-time interpretation of mathematical expressions tommy.thorn@gmail.com (Tommy Thorn) (2006-11-13) |

Re: Fastening the run-time interpretation of mathematical expressions 148f3wg02@sneakemail.com (Karsten Nyblad) (2006-11-15) |

Re: Fastening the run-time interpretation of mathematical expressions martin@gkc.org.uk (Martin Ward) (2006-11-15) |

[3 later articles] |

From: | haberg@math.su.se (Hans Aberg) |

Newsgroups: | comp.compilers |

Date: | 11 Nov 2006 21:30:49 -0500 |

Organization: | Mathematics |

References: | 06-11-052 |

Keywords: | interpreter |

"PAolo" <paolopantaleo@gmail.com> wrote:

*> I am writing a piece of software that accept in input the definition*

*> of a mathematical function and a list of points in which evaluate the*

*> function and puts the result in output. The main feature of the progam*

*> is that once the function definition is read, the program evaluates*

*> the function very fast [Well this isn't done already...]. I was*

*> thinking to build a pair of stacks, one containig the operands*

*> (floats) and the other containig the operators (function pointers).*

This is typically used when the user is allowed to define operators

with a large number of precedence levels. One can then integrate it in

an otherwise static grammar. I once write it, but I do not immediately

recall if I used one or two stacks. The trick is to the operator

precedence. Add an end marker # to the expression; when it appears,

the rest of the stacks are computed. Try a two value/operator stack

approach on "a + b * c #". First push 'a', '+', 'b'. When the '*'

appears, check the operator stack, to see that it is a '+' with lower

precedence on top; so then '*' and 'c' should be pushed. Then '#'

appears, saying that all stuff on the stacks should be computed, first

b * c, which is put back on the value stack, and then '+' is applied

to that.

When you know how to compute the expression, converting it to other

formats, such as the Australian notation

<http://en.wikipedia.org/wiki/Reverse_Polish_notation>, is easy.

--

Hans Aberg

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