I think this is the answer: >> type? first [/] == word! >> type? first [/a] == refinement!
No, it don't evaluate the values but Yes it needs to separate the values. A block! is a series of any-word!, and it put a blank between values. ... I think. > > Hi Volker, > > Thanks for your answer... (see comment at the bottom) > > > > > > > >> f: func [arg][probe arg] > > > >> f [/a/b/c] > > > [/a /b /c] > > > > > > So there's some kind of processing of the block content? I thought=20 > > > block protected their content from evaluation ? How comes ?=20 > > Any idea ? > > > > >=20 > > Syntax-error. you wrote /a/b/c , not a/b/c . rebol does not=20 > > look for a space here, so it takes /a, then /b, then /c . And=20 > > molds it back with spaces. weird parser. > =20 > The notation /a/b/c is what I need, that's why I enclose it into a > block, and not pass it as a path or a lit-path.=20 > But the question is: why does the evaluator evaluate the block, while I > didn't tell him to do so. Block! should protect my data from evaluation > ! Am I wrong about that ? -- To unsubscribe from the list, just send an email to lists at rebol.com with unsubscribe as the subject.
