|Bill Allombert on Tue, 23 Feb 2010 14:18:40 +0100|
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|Re: Getting the variable number of a lexical local variable|
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.