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Re: understanding code
*To*: mathgroup at smc.vnet.net
*Subject*: [mg114219] Re: understanding code
*From*: Leonid Shifrin <lshifr at gmail.com>
*Date*: Sat, 27 Nov 2010 03:38:21 -0500 (EST)
Hi Bill,
PreIncrement must have been implemented on the higher level, using Set,
which
can be illustrated as follows:
In[12]:=
i[1]=0
Out[12]= 0
In[13]:= On[Set]
In[14]:= i[1]++
During evaluation of In[14]:= Set::trace: i[1]=1 --> 1. >>
Out[14]= 0
In[15]:= Off[Set]
I would not be surprised in the implementation looks something like this:
SetAttributes[preIncrement, HoldFirst];
preIncrement[expr_]:=
With[{value = expr}, expr=value+1;value];
In any case, there seem to be no additional restrictions imposed by
PreIncrement
on an expression being incremented, as compared to those of Set (that it
must be an
L - value). Particularly, indexed variables like i[1] are ok.
Regards,
Leonid
On Fri, Nov 26, 2010 at 1:26 PM, Bill Rowe <readnews at sbcglobal.net> wrote:
> On 11/25/10 at 5:57 AM, sam.takoy at yahoo.com (Sam Takoy) wrote:
>
> >The following code:
>
> >i[_] = 0;
> >i[1]++
> >i[1]++
> >i[1]++
> >i[1]++
>
> >return 0 1 2 3 4. I'd like to understand whats going on here. That
> >is,
>
> >What is i? Is it a function? And what is i[1]? Is it a function or
> >a value? Etc...
>
> It is a function. i[1] is the function i evaluated at i. You can
> see what is going on by using Trace, i.e.,
>
> In[3]:= Trace[i[1]++]
>
> Out[3]= {i(1)++,{i(1),0},{i(1)=1,1},0}
>
> As you can see, i[1] gets evaluated to whatever value it had
> previously which initially is zero. Then the result gets
> incremented by 1 and is then assigned to be the new value for
> i[1]. This last step is not something I would have expected
> without your example and I don't know why it occurs.
>
> I can see times where this behavior might be useful. For
> example, it is at times convenient to see i[1] as a subscripted
> variable. Thinking of i[1] this way, the notation i[1]++ would
> be interpreted as incrementing the subscripted variable by one
> and the Mathematica behavior is consistent with this.
>
>
>
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