Re: Getting the variable number of a lexical local variable

[Charles Greathouse <charles.greathouse@case.edu>]

> Actually, it is possible: just use the code 'n', not 'V='.

I mean that it's not (apparently) possible to get it from the existing
signature.

> Well, you can use
> sumformal(x,1, n, x^3 + 3)
> instead. The only difference is the use of a comma instead of =.

I guess that's not too bad.  I don't suppose there's a way to overload
it so that I can have it called with V=GGG or nGGG?

Charles Greathouse
Analyst/Programmer
Case Western Reserve University

On Tue, Feb 23, 2010 at 8:11 AM, Bill Allombert
<Bill.Allombert@math.u-bordeaux1.fr> wrote:
> On Mon, Feb 22, 2010 at 04:08:18PM -0500, Charles Greathouse wrote:
>> I see, thanks for the explanation.  So it's not possible to know what
>> variable number is used;
>
> Actually, it is possible: just use the code 'n', not 'V='.
>
>> maybe I should just change the format to
>> sumformal(1, n, x^3 + 3)
>> rather than
>> sumformal(x=1, n, x^3 + 3)
>> since I don't want to make users think they can control what variable is used.
>
> Well, you can use
> sumformal(x,1, n, x^3 + 3)
> instead. The only difference is the use of a comma instead of =.
>
>> But wait:
>> x=10;y=100;for(x=1,2,print(x^2*y))
>> and
>> x=10;y=100;for(y=1,2,print(x^2*y))
>> give different results.  How can they tell?
>
> They do not: they simply use the code 'I' to evaluate the expression
> 'print(x^2*y)' while setting the loop variable to 1 and 2. They do not
> know what is the name of the loop variable, and there is no polynomial
> variables associated to that name. And in any case 'print(x^2*y)' is
> not a polynomial expression.
>
> Cheers,
> Bill.