Wed, 16 Dec 1992 20:04:36 GMT

Related articles |
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[2 earlier articles] |

Re: question on control dependence cliffc@rice.edu (1992-12-15) |

Re: question on control dependence preston@dawn.cs.rice.edu (1992-12-15) |

Re: question on control dependence preston@dawn.cs.rice.edu (1992-12-15) |

Re: question on control dependence paco@cs.rice.edu (Paul Havlak) (1992-12-15) |

Re: question on control dependence bwilson@shasta.stanford.edu (1992-12-15) |

Re: question on control dependence paco@cs.rice.edu (1992-12-16) |

Re: question on control dependence dkchen@sp91.csrd.uiuc.edu (1992-12-16) |

Newsgroups: | comp.compilers |

From: | dkchen@sp91.csrd.uiuc.edu (Ding-Kai Chen) |

Organization: | UIUC Center for Supercomputing Research and Development |

Date: | Wed, 16 Dec 1992 20:04:36 GMT |

References: | 92-12-056 92-12-070 |

Keywords: | design |

preston@dawn.cs.rice.edu (Preston Briggs) writes:

*>The general case of more than 2 succesoors is handled like this...*

*> CD(X) = union(P(S)) - intersection(P(S)), for S in successors(X)*

*>(This is a much simpler version of another answer I posted earlier)*

This can be simplified (also with less computation) further:

CD(X) = union(P(S)) - P(X)

I think it is not difficut to prove that

{intersection(P(S)), for S in successors(X)}=P(X)

Similarly,

CD(X,L) = P(X_L) - P(X)

where CD(X,L) is the set of nodes control depend on X with label L and X_L

is the successor of X with label L.

Ding-Kai Chen

--

Ding-Kai Chen University of Illinois

(217)244-0046 Center for Supercomputing R&D,

dkchen@csrd.uiuc.edu 465 CSRL 1308 W. Main St. Urbana, IL 61801

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