What does FullForm[ ] actually do?
- To: mathgroup at smc.vnet.net
- Subject: [mg90467] What does FullForm[ ] actually do?
- From: AES <siegman at stanford.edu>
- Date: Thu, 10 Jul 2008 06:35:26 -0400 (EDT)
- Organization: Stanford University
I'm having trouble grasping just what it is that FullForm[ ] does. For example: 1) Page 424 of the M5 Book says: "FullForm[expr] shows the internal form of an expression in explicit functional notation" and gives a moderately complex example of this. I don't find it clearly stated, however, there or anywhere else in the immediately accessible M5 or M6 documentation, whether executing "FullForm[expr]" also executes the "expr" itself, although some experimenting seems to show that it does. If so, perhaps the documentation ought to make this clear. . . ? 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. But isn't the full expr that's operated on by FullForm, or that forms the argument of FullForm[expr], the complete expression "y = {a+b}". That's what fits the stated syntax of the FullForm[expr] command -- and moreover, page 230 of the M5 Book says: 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"? 3) Even without the "y =" part, suppose we execute either of the inputs FullForm[y={a+b};] or FullForm[{a+b};] (with a semicolon added). Either of these gives us back the displayed result "Null" -- but the first of them also sets the value of y to {a+b}, confirming that the complete expression inside the FullForm brackets was indeed executed. 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? 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." 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 . . . ?
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