emacs的對齊功能參數垂直

Question

我試圖實現以下目標：

void foo(int one, 
int const & two, 
Bar three) 

到

void foo(int   one, 
      inst const & two, 
      Bar   three) 

這可能使用的align-regex函數來完成（有或沒有前綴）？

更一般地說，正則表達式中的分組意味着什麼（它是否被認爲是'列'）？那麼「括號內的小組需要修改（證明是否否定）」？

由於

Answer 1

參見C-H˚Falign-regexpRET和，特別地，鏈接C-Hvalign-rules-listRET提供了一些用於對準的最佳文檔。

「要修改的組」表示在對齊時將縮小或擴大的模式中的組。你幾乎總是希望這個小組純粹是空白的，以避免刪除實際內容。

的組參數 - 交互「括號組來修改（如果證明負）」 - 是在正則表達式問題的組的從1

的理由部分計數的數目，是有一點麻煩。如果你提供一個負數，則同一組作爲如果該號碼是積極的，但align-rules-list變量的「證明」的行爲也引發：

`justify' 
It is possible with `regexp' and `group' to identify a 
character group that contains more than just whitespace 
characters. By default, any non-whitespace characters in 
that group will also be deleted while aligning the 
alignment character. However, if the `justify' attribute 
is set to a non-nil value, only the initial whitespace 
characters within that group will be deleted. This has 
the effect of right-justifying the characters that remain, 
and can be used for outdenting or just plain old right- 
justification. 

Answer 2

IMO這是一個情況下使用正則表達式的變得複雜。使用語法-ppss的功能更容易：

(defun my-arguments-indent() 
    "When called from inside an arguments list, indent it. " 
    (interactive "*") 
    (save-excursion 
    (let* ((pps (syntax-ppss)) 
      (orig (point)) 
      indent) 
     (while (and (nth 1 pps)(not (eobp))) 
     (setq indent (save-excursion 
         (when (nth 1 pps) 
         (goto-char (nth 1 pps)) 
         (forward-char 1) 
         (skip-chars-forward " \t") 
         (current-column)))) 
     (when (and (< orig (line-beginning-position)) indent) 
      (beginning-of-line) 
      (fixup-whitespace) 
      (indent-to indent)) 
     (forward-line 1) 
     (back-to-indentation) 
     (setq pps (syntax-ppss))))))