elisp：有没有办法获取当前.el模块的名称（如C中的__FILE__）？ [英] elisp: Is there a way to get the name of the current .el module (like __FILE__ in C)?

问题描述

 （message（concatLoading （expand-file-name（current-elisp-module）。））））
  

解决方案

您可以使用变量 load-file-name ，由 load ，记录如下：

由`load'加载的文件的全名。

如手册中所详述：

当Emacs正在加载一个文件，该变量的值是该文件的名称，正如Emacs在之前描述的搜索过程中发现的那样在这个部分。

注意： buffer-file-name 作为例程不起作用正如你可能期望的那样。

At the top of my elisp module, I want to do something as simple as:

(message (concat "Loading " (expand-file-name (current-elisp-module) ".")))

解决方案

You can use the variable load-file-name, which is set by the function load, documented as follows:

Full name of file being loaded by `load'.

As elaborated in the manual:

When Emacs is in the process of loading a file, this variable’s value is the name of that file, as Emacs found it during the search described earlier in this section.

Note: buffer-file-name as a routine does not work as you might expect it to.

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