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Re: What does FullForm[ ] actually do?

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  • Subject: [mg90520] Re: [mg90467] What does FullForm[ ] actually do?
  • From: "W_Craig Carter" <ccarter at>
  • Date: Fri, 11 Jul 2008 02:06:19 -0400 (EDT)
  • References: <>

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 . . . ?


W. Craig Carter

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