Acad.lin file load by lisp

Pl. can you give me a lisp code to load acad.lin file

What exactly is the goal here? Is it to load every linetype that is included in acad.lin?

Something like this is what I have in my old notes.

(command "_.-linetype" "load" "*" "path to .lin" "") 

You can use the command line version of the Layer command (-LAYER) to load linetypes as well, at least you used to and seems to work in AutoCAD 2011.

Try this.

(defun c:loadltypes ( / fn f rl ltname ltypes )
 (setq fn (open (setq f (findfile "acadiso.lin")) "r"))
 (while (setq rl (read-line fn))
   (if (eq "*" (substr rl 1 1)) (setq ltname (substr rl 1 (vl-string-position (ascii ",") rl))))
   (if ltname
     (progn
       (setq ltname (vl-string-left-trim "*" ltname))
       (setq ltypes (cons ltname ltypes))
     )
   )
 )
 (foreach ln ltypes
   (command "_.linetype" "l" f "" "s" ln "")
 )
 (command "_.linetype" "s" "ByLayer" "")
 (princ)
)

Lee Mac's http://www.lee-mac.com/loadlinetype.html will search lin files and load a list of linetypes giving the option to redefine them. I use several lin files as an Arial font used by an Arial.lin produces clean visible dotted and arrow linetypes. I use another lisp to add the text style first if needed.

*STriangle, Solid Triangle__/_\_/_\__/_\_/_\_/_\_
S,.000000001,-.05,["\U+25BA",Arial,S=0.02,R=90,Y=-.0019],-.05
*ARROWH,Arrowhead  >  >  >  >  >  >  >
A,.000001,-3,["\U+25BA",Arial,S=3,R=0,X=-1.35,Y=-1.35],.000001,-4
*ArialARROW,Arrow1  >  >  >  >  >  >  >
S,.000001,-3,["\U+25AC",Arial,S=3,R=0,X=-1.35,Y=-1.05],.000001,["\U+25BA",Arial,S=3,R=0,X=0.5,Y=-1.42],.000001,-4
*ArialDOT,Dot . . . . . . . . . . . . . . . . . . . . . . . .
A,0,-.125,["\U+2022",Arial,S=.05,Y=-.024],-.125
*ArialDOT2,Dot (.5x) ........................................
A,0,-.0625,["\U+2022",Arial,S=.05,Y=-.012],-.0625
*ArialDOTX2,Dot (2x) .  .  .  .  .  .  .  .  .  .  .  .  .  .
A,0,-.25,["\U+2022",Arial,S=.05,Y=-.048],-.25

My Acad.lin and FDOT.lin files are huge. I'd never load all of either of them in a drawing.

Thanks everybody.