Subject: 28.0.50; calculator.el: Cannot input negative exponents

From: Chris Zheng <[email-masked]>
Date: Sun, 17 May 2020 13:53:12 +0800

Hello,

With current master, starting Emacs with `-Q`:

1. M-x calculator
2. Input 1e-3 RET

You will get 1 instead of 0.001. I believe this happens since Emacs 26.1. The root cause is the `calculator-string-to-number` function in lisp/calculator.el. Now the function gives

(calculator-string-to-number "1e-3") => 1.0
(calculator-string-to-number "1e3") => 1000.0
(calculator-string-to-number "1e") => 1.0

The funcional code is 

(replace-regexp-in-string
                  "[eE][+-]?\\([^0-9].*\\)?$" "e0\\1" str)

It changes `1e-3` to `1e0-3` that is recognized as 1. I have a possible fix attached below. Please see if it is correct. 

Thank you,

Chris Zheng

 From 7693d7072e4787c4b0663f490be5d83c1d9a6ee3 Mon Sep 17 00:00:00 2001
From: Chris Zheng <[email-masked]>
Date: Sun, 17 May 2020 13:43:35 +0800
Subject: [PATCH 1/1] Fix calculator exponent input

* lisp/calculator.el (calculator-string-to-number): Change the regexp
to correctly deal with negative exponents.
---
  lisp/calculator.el | 2 +-
  1 file changed, 1 insertion(+), 1 deletion(-)

diff --git a/lisp/calculator.el b/lisp/calculator.el
index 7e0b2fcc6a..fa3eb19099 100644
--- a/lisp/calculator.el
+++ b/lisp/calculator.el
@@ -863,7 +863,7 @@ calculator-string-to-number
      (let* ((str (replace-regexp-in-string
                   "\\.\\([^0-9].*\\)?$" ".0\\1" str))
             (str (replace-regexp-in-string
-                 "[eE][+-]?\\([^0-9].*\\)?$" "e0\\1" str)))
+                 "[eE]\\([+-]?\\)?$" "e\\10" str)))
        (float (string-to-number str)))))
  
  (defun calculator-push-curnum ()
-- 
2.16.1.windows.1