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MathGroup Archive 2000

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Re: With[{software=Mathematica}, Frustration]

  • To: mathgroup at smc.vnet.net
  • Subject: [mg24370] Re: With[{software=Mathematica}, Frustration]
  • From: "Drago Ganic" <drago.ganic at in2.hr>
  • Date: Wed, 12 Jul 2000 23:13:11 -0400 (EDT)
  • References: <8k3of0$428@smc.vnet.net>
  • Sender: owner-wri-mathgroup at wolfram.com

Hi,

well in my practical experience with Mathematica (a Year of hobby learning) the subject to this post you have choose is quite correct. The only suggestion I can give You is to buy a lot of books (I would like to know how many non-genious people have learned the deep basics of Mathematica only from the Mathematica book).

Doing so I can give you the answer:

In[6]:=Attributes[With]
Out[6]= {HoldAll, Protected}

In[8]:= Attributes[ReplaceAll]
Out[8]={Protected}

As you can see "With" does not evaluate c,  therefore {a->2, b->3} acts on c, and not on a*b (like  in "ReplaceAll").

If you want that "With" behaves the same as "ReplaceAll" you schould use "Evaluate" ....

In[10]:= With[{a = 2, b = 3}, Evaluate[c]]
Out[10]=6

or if you want that "ReplaceAll" behaves the same as "With" you schould use "Unevaluated" 

In[5]:= Unevaluated[c] /. {a -> 2, b -> 3}
Out[5]= a b


as Stephen said in The Book (you have not quooted the * ... * part ):

*Except for the question of when x and body are evaluated*, With[ax = aa, body] works essentially like body /. x -> a. 


So, read carefully.

Greeting from Croatia,
Drago Ganic


"AES" <siegman at stanford.edu> wrote in message news:8k3of0$428 at smc.vnet.net...
> Pages 359-360 of The Mathematica Book says (admittedly, taken a little 
> out of context), 
> 
>    "You can think of  With  as a generalization of the /. operator. . ."
> 
> and
> 
>    " With[{x=x0}, body]  works essentially like body /. x->x0 . . . "
> 
> Great, looks neat, let's try it for evaluating expressions without 
> permanently setting the variables in them:
> 
>    In[1]:= c = a b
> 
>    Out[1]= a b
> 
>    In[2]:= c
> 
>    Out[2]= a b
> 
>    In[10]:= c /. {a -> 2, b -> 3}
> 
>    Out[10]= 6
> 
>    In[3]:= With[{a = 2, b = 3}, c]
> 
>    Out[3]= a b
> 
> *Not* what I was hoping for  . . .
> 



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