       Re: ClearAll[f]; f[x_] := x^2; f[y_] :=y^4; (*What is:*) f

• To: mathgroup at smc.vnet.net
• Subject: [mg84387] Re: ClearAll[f]; f[x_] := x^2; f[y_] :=y^4; (*What is:*) f
• From: dh <dh at metrohm.ch>
• Date: Thu, 20 Dec 2007 16:52:06 -0500 (EST)
• References: <fkcueh\$5d9\$1@smc.vnet.net>

```
Hi,

with "Information[..] or ?? for short" you can see what is stored. In

version 6, as you can convince yourself, only one definition ist stored.

I do not have version 2 or 3 available so I can not tell what is going

on there.

Daniel

cebailey wrote:

> ClearAll[f]; f[x_] := x^2; f[y_] :=y^4; (*What is:*) f

>

> Evaluating this line in Mathematica 5.2 or Mathematica 6 returns 16. This makes sense, because the second definition replaces the first, as we can see when ?f returns:

> Global`f

> f[y_]:=y^4

>

> But in _A_Physicist's_Guide_to_Mathematica_ on p.314, Patrick Tam shows an example like this returning the other answer, 4, defined in the first definition. He then demonstrates that ?f returns:

> Global`f

> f[x_] := x^2

> f[y_]:= y^4

> He says his book was developed with Mathematica 2.2 and a prerelease of Mathematica 3 and is compatible with both.

>

> He goes on to explain:

> "Contrary to expectation, Mathematica used the first definition. The ? operator reveals that Mathematica stores both definitions in the global rule base, giving higher priority to the first definition. (This problem cannot, perhaps, be called a bug because developers of Mathematica are well aware of this design flaw, which is quite difficult to mend....)"

>

> What is he talking about? Did Mathematica  2.2 and 3 treat this differently? If earlier versions worked in this surprising way, there must have been a reason - what was it? Was it changed to prevent surprises like this example? Did changing it create other unfortunate consequences? Was Tam just wrong? Or do I misunderstand?

>

```

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