From a0cf5a607c1e5db79803473b1856f5c5c7636c66 Mon Sep 17 00:00:00 2001 From: Raymond Toy Date: Wed, 26 Jun 2024 05:27:12 -0700 Subject: [PATCH] Rename *ll* and *ul* to ll and ul in ratfnt The function `ratfnt` doesn't really need the special variables since they're arguments, so rename the args to `ll` and `ul` to break that connection. Full testsuite + share passes with cmucl and ccl64. --- src/defint.lisp | 16 ++++++++-------- 1 file changed, 8 insertions(+), 8 deletions(-) diff --git a/src/defint.lisp b/src/defint.lisp index 8169d9c5c..3f9c66409 100644 --- a/src/defint.lisp +++ b/src/defint.lisp @@ -679,7 +679,7 @@ in the interval of integration.") ;; Integrate rational functions over a finite interval by doing the ;; polynomial part directly, and converting the rational part to an ;; integral from 0 to inf. This is evaluated via residues. -(defun ratfnt (exp ivar *ll* *ul*) +(defun ratfnt (exp ivar ll ul) (let ((e (pqr exp ivar))) ;; PQR divides the rational expression and returns the quotient ;; and remainder @@ -690,23 +690,23 @@ in the interval of integration.") (cond ((equal 0. (car e)) ;; No polynomial part - (let ((ans (try-antideriv exp *ll* *ul*))) + (let ((ans (try-antideriv exp ll ul))) (if ans ans - (cv exp ivar *ll* *ul*)))) + (cv exp ivar ll ul)))) ((equal 0. (cdr e)) ;; Only polynomial part - (eezz (car e) *ll* *ul* ivar)) + (eezz (car e) ll ul ivar)) (t ;; A non-zero quotient and remainder. Combine the results ;; together. - (let ((ans (try-antideriv (m// (cdr e) dn*) *ll* *ul*))) + (let ((ans (try-antideriv (m// (cdr e) dn*) ll ul))) (cond (ans - (m+t (eezz (car e) *ll* *ul* ivar) + (m+t (eezz (car e) ll ul ivar) ans)) (t - (m+t (eezz (car e) *ll* *ul* ivar) - (cv (m// (cdr e) dn*) ivar *ll* *ul*)))))))))) + (m+t (eezz (car e) ll ul ivar) + (cv (m// (cdr e) dn*) ivar ll ul)))))))))) ;; I think this takes a rational expression E, and finds the ;; polynomial part. A cons is returned. The car is the quotient and -- 2.11.4.GIT