Re: What does FullForm[ ] actually do?

*To*: mathgroup at smc.vnet.net*Subject*: [mg90520] Re: [mg90467] What does FullForm[ ] actually do?*From*: "W_Craig Carter" <ccarter at mit.edu>*Date*: Fri, 11 Jul 2008 02:06:19 -0400 (EDT)*References*: <200807101035.GAA15381@smc.vnet.net>

Hello AES, I believe this example demonstrates what is going on: innermost expressions are evaluated prior to the "FullForm" FullForm[Hold[y = {a + b};]] ReleaseHold[FullForm[Hold[y = {a + b};]]] > 2) Then, just as an example, executing FullForm[y = {a+b}] gives (that > is, Prints, or displays on screen) only the output List[a,b] -- the "y > =" is gone. > > 2.1.1 Everything Is an Expression > > "...everything you type into M is treated as an expression." > > Isn't the "y = " part typed in as part of expr also? Doesn't it have to > have some form of "internal form"? I believe it is and does (Set in the above example) > But what is it that's "null" here? The "{a+b}" or even the "{a+b};" > part clearly isn't null, since it gets put into y in the first form. > And isn't ";" (the semicolon) also an expression, and also part of the > "{a+b};" expression? ("Everything in M is an expression.") Don't both > ; and {a+b}; have to have an internal form? Null is the result of the last expression in the CompoundExpression, which was nothing. > > The bottom line seems to be that FullForm[{a+b}], when executed, in some > cases does as claimed "show the internal form of that expression in > explicit functional notation." > This is true if we interpret what is being digested by FullForm. Hold causes indigestion. > However, the FullForm[{a+b};] or FullForm[y={a+b};] examples seem to > show that executing FullForm[expr] returns the result of executing that > expr, not the expression itself . . . ? Exactly. > > -- W. Craig Carter

**References**:**What does FullForm[ ] actually do?***From:*AES <siegman@stanford.edu>