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I would like create function that ask me how length I want the Pomodoro but it say wrong type argument. 

(defun changePomoLength ()
  (interactive)
  (set-variable org-pomodoro-length (read-string "How longth: "))
  )

When I open set-variable function description. it signature is

(set-variable VARIABLE VALUE &optional MAKE-LOCAL)

VALUE is a LISP Object, is string read from (read-string) function not count as LISP object? how to make my string became LISP object?

Drew
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moanrisy
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1 Answers1

3

VARIABLE is a variable, that is, a symbol.

Instead of passing a symbol as the first argument, you passed its value.

set-variable is a function, not a macro or special form. It first evaluates each of its arguments, then acts on their values. The first argument you passed is org-pomodoro-length. That is, because the args get evaluated, you passed the value of org-pomodoro-length - which apparently is not a symbol. Hence the error.

This is probably what you wanted - note the quote mark:

(defun changePomoLength ()
  (interactive)
  (set-variable 'org-pomodoro-length (read-string "How longth: ")))
Drew
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  • this solution work (I try enter 15) but I got another error when running orrg-pomdoro.

    user-error: Value ‘"15"’ does not match type integer of org-pomodoro-length

    is this mean that I should not save the variable as string? can you help me again how to save org-pomdoro-length as integer?

     – moanrisy Apr 21 '20 at 11:56
  • Nevermind, I change (read-string) to (read) and it work, thank you very much. – moanrisy Apr 21 '20 at 12:01