From: SourceForge.net <no...@so...> - 2005-03-23 20:22:45
|
Bugs item #1169382, was opened at 2005-03-23 12:22 Message generated for change (Tracker Item Submitted) made by Item Submitter You can respond by visiting: https://sourceforge.net/tracker/?func=detail&atid=104933&aid=1169382&group_id=4933 Category: None Group: None Status: Open Resolution: None Priority: 5 Submitted By: Nobody/Anonymous (nobody) Assigned to: Nobody/Anonymous (nobody) Summary: for matrix A, A[1,0] give "[" Initial Comment: Example : (%i1) A:matrix([1,2],[3,4]); [1 2] (%o1) [ ] [ 3 4] (%i2) A[1,0]; (%o2) [ ---------------------------------------------------------------------- You can respond by visiting: https://sourceforge.net/tracker/?func=detail&atid=104933&aid=1169382&group_id=4933 |
From: SourceForge.net <no...@so...> - 2005-03-24 04:08:11
|
Bugs item #1169382, was opened at 2005-03-23 14:22 Message generated for change (Comment added) made by willisbl You can respond by visiting: https://sourceforge.net/tracker/?func=detail&atid=104933&aid=1169382&group_id=4933 Category: None Group: None Status: Open Resolution: None Priority: 5 Submitted By: Nobody/Anonymous (nobody) Assigned to: Nobody/Anonymous (nobody) Summary: for matrix A, A[1,0] give "[" Initial Comment: Example : (%i1) A:matrix([1,2],[3,4]); [1 2] (%o1) [ ] [ 3 4] (%i2) A[1,0]; (%o2) [ ---------------------------------------------------------------------- >Comment By: Barton Willis (willisbl) Date: 2005-03-23 22:08 Message: Logged In: YES user_id=895922 Matrices and lists are indexed starting at 1. If you want the element of A in the 2nd row first column, use A[2,1]. So why does A[1,0] evaluate to [ ? Because A[1,0] == part(A,1,0). The zeroth part of a non-atom is its operator. Now, part(A,1) is a list and part(A,1,0) is the operator of a list. Finally, the operator of a list is [. I don't consider A[1,0] --> [ to be a bug. Others may have different opinions. Barton ---------------------------------------------------------------------- You can respond by visiting: https://sourceforge.net/tracker/?func=detail&atid=104933&aid=1169382&group_id=4933 |
From: SourceForge.net <no...@so...> - 2006-08-12 14:34:06
|
Bugs item #1169382, was opened at 2005-03-23 13:22 Message generated for change (Comment added) made by robert_dodier You can respond by visiting: https://sourceforge.net/tracker/?func=detail&atid=104933&aid=1169382&group_id=4933 Please note that this message will contain a full copy of the comment thread, including the initial issue submission, for this request, not just the latest update. >Category: Lisp Core Group: None >Status: Closed >Resolution: Fixed Priority: 5 Submitted By: Nobody/Anonymous (nobody) Assigned to: Nobody/Anonymous (nobody) Summary: for matrix A, A[1,0] give "[" Initial Comment: Example : (%i1) A:matrix([1,2],[3,4]); [1 2] (%o1) [ ] [ 3 4] (%i2) A[1,0]; (%o2) [ ---------------------------------------------------------------------- >Comment By: Robert Dodier (robert_dodier) Date: 2006-08-12 08:34 Message: Logged In: YES user_id=501686 Fixed by r1.27 src/mlisp.lisp -- now 0 index => "No such matrix element". See also bug report 705734 (list element 0 => "["). ---------------------------------------------------------------------- Comment By: Barton Willis (willisbl) Date: 2005-03-23 21:08 Message: Logged In: YES user_id=895922 Matrices and lists are indexed starting at 1. If you want the element of A in the 2nd row first column, use A[2,1]. So why does A[1,0] evaluate to [ ? Because A[1,0] == part(A,1,0). The zeroth part of a non-atom is its operator. Now, part(A,1) is a list and part(A,1,0) is the operator of a list. Finally, the operator of a list is [. I don't consider A[1,0] --> [ to be a bug. Others may have different opinions. Barton ---------------------------------------------------------------------- You can respond by visiting: https://sourceforge.net/tracker/?func=detail&atid=104933&aid=1169382&group_id=4933 |