[英]How to call a function in an elisp program, and follow the interactive path?
If I call ispell-check-version
interactively (through Mx ) it got this answer: 如果我以交互方式调用
ispell-check-version
(通过Mx ),它得到了这个答案:
@(#) International Ispell Version 3.3.02 12 Jun 2005, ispell.el 3.6 - 7-Jan-2003
@(#)International Ispell Version 3.3.02 2005年6月12日,ispell.el 3。6 - 7-Jan-2003
Fine. 精细。 Now let's say I want to call it from a program.
现在让我们说我想从程序中调用它。 I expect to get the same result.
我希望得到相同的结果。 But the function behaves differently when not call interactively.
但是当不以交互方式调用时,函数的行为会有所不同。 Its documentation tells so, and indeed it does.
它的文档说明了,事实确实如此。 Let's try in the
*scratch*
buffer (evaluated with Cu Cx Ce ): 让我们尝试
*scratch*
缓冲区(用Cu Cx Ce评估):
(ispell-check-version)
returns me : "/opt/local/lib" !! 给我回复:“/ opt / local / lib”!! Not exepected.. How can I make the function behaves and the return the same result now that I call it "programatically" as when I was calling it "interactively" please?
没有被考虑..如果我以“编程方式”称它为“交互式地”,请如何使函数表现并返回相同的结果?
call-interactively
是您正在寻找的。
(call-interactively 'ispell-check-version)
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