Am 01.04.2010 04:31, schrieb Ted:
Hello,
I'm hoping to call a "static" dictionary function using call(). By
"static" I mean that the keyword 'dict' is not used in the function's
definition. I use this nomenclature in the hopes that the effect of
this keyword is to declare a static member function as is possible in
java/C++/etc, ie to restrict the scope of the function name without
requiring it to be associated with an object. However this doesn't
seem to work. For example:
" Setup:
let testdict = { }
funct! testdict.funct()
echo "called"
endfunct
" Tests:
" Following each line is an indented comment containing its
" output in message land, ie what was echoed.
call testdict.funct()
" called
echo testdict.funct
" 667
echo string(testdict.funct)
" function('667')
echo function('667')
" E475: Invalid argument: 667
echo function('testdict.funct')
" testdict.funct
call call(testdict.funct, [ ])
" E725: Calling dict function without Dictionary: 667
If it wants a dict, then give it a dict:
call call(testdict.funct, [], {})
" Same deal if there's an intermediate variable involved.
let TestdictDotFunct = testdict.funct
echo TestdictDotFunct
" 667
echo string(TestdictDotFunct)
" function('667')
call TestdictDotFunct()
" E725: Calling dict function without Dictionary: 667
call call(TestdictDotFunct, [], {})
From the help topic *E725*:
It is also possible to add a function without the "dict" attribute
as a Funcref to a Dictionary, but the "self" variable is not
available then.
Applies when the function is global:
func! GlobalFunc()
echo exists("self")
endfunc
let testdict.onemorefunc = function("GlobalFunc")
So logic would seem to indicate that if "self" is not available, then
it should be possible to call the function referenced by the Funcref
without a Dictionary. However this doesn't seem to be the case. Am I
missing something?
It's because the following implies the dict attribute:
let testdict = { }
funct! testdict.funct()
...
HTH (no warranty),
Andy
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