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Re: The side-effects of mixing TraditionalForm inside expressions.
On 7 Jul 2010, at 20:42, AES wrote: > > Help Browser says: > > Unevaluated[expr] > represents the unevaluated form of expr when it appears as > the argument to a function. > > So, what does Unevaluated[expr] do when it is NOT the argument to a > function? It does nothing. You could have checked it yourself: In:== Unevaluated Out== Unevaluated > > (The previous statement I'd say clearly implies that there are times > when Unevaluated[expr] does appear as an arg to function, and therefore > times when it does not -- and both need to be defined.) > > So, if Unevaluated[expr] is entered all by itself in a cell, is it then > "the argument to a function"? No. > > If the cell contains 1 + Unevaluated[expr] as above, does that make the > Unevaluated[expr] an argument to a function -- or not? 1+Unevaluated is actually Plus[1,Unevaluated] so Unevaluated1] is here an argument to a function. Also here In:== Identity[Unevaluated] Out== 1 Unevaluated is again an argument to an explicit function. In my opinion the documentation is perfectly clear and correct in this particular case. Of course it is always possible to misunderstand or fail to understand anything, however clear, if one tries hard enough. Andrzej Kozlowski