Hello! On Thu, 2007-05-24 at 14:50 -0400, Warren Weckesser wrote:

This works for me:

string s = ex_to<symbol>(E).get_name();

Firstly, I'd like to note that GiNaC expressions do not have any "names" at all. Only symbols have "names" which serve merely as a printing representation. Two different symbols may have the same name, as in symbol x("x"), y("x"); // valid code or namespace A { symbol x("x"); } namespace B { symbol x("x"); } So your code will work if (and only if) E is a symbol. And it is good idea to check if E is indeed symbol, since ex_to<T> functions are unsafe (as documented in the manual), e.g. string s = "dunno"; if (is_a<symbol>(E)) s = ex_to<symbol>(E).get_name(); On Fri, May 25, 2007 at 10:48:44AM +0200, Javier Ros Ganuza wrote:

I have a similar problem, if I define the clifford units this way,

varidx nu(symbol("nu", "\\nu"), 3); ex basis1 = clifford_unit(mu, diag_matrix(lst(1, 1, 1)),0);

ex e00 = basis1.subs(mu == 0), e01 = basis1.subs(mu == 1), e02 = basis1.subs(mu == 2);

Then they are ploted

cout << e00 << endl;

Like this

e~0 I want to rename using symbol.set_name(string), but when I do

cout << ex_to<symbol>(e00).get_name() << endl;

This should segfault. e00 is not a symbol, it is element of Clifford algebra. So ex_to<symbol> is meaningless here...

Anything gets plotted.

Well, some sunny day it may even print something. Probably you are lucky :)

So I suspect that e00 is not a symbol an then,

It is definitely not a symbol. It is an element of Clifford algebra.

It woul be interesting to have this posibility, tough elements of diferent basis are right now plotted in the same way.

I don't think renaming symbols will help here.

Basis1 e00, plots as e~0

Basis2 e10, plots as e~0

Nevertheless if plotting in latex style, units are plotted in a different way.

Indeed. And this is $ cat test.cpp #include <iostream> #include <ginac/ginac.h> using namespace std; using namespace GiNaC; int main(int argc, char** argv) { varidx nu(symbol("nu", "\\nu"), 3); ex basis1 = clifford_unit(nu, diag_matrix(lst(1, 1, 1)), 0); ex basis2 = clifford_unit(nu, diag_matrix(lst(3, 1, 2)), 1); ex e10 = basis1.subs(nu == 0), e20 = basis2.subs(nu == 1); cout << latex << "e10 = " << e10 << endl << "e20 = " << e20 << endl << dflt << "e10 = " << e10 << endl << "e20 = " << e20 << endl; return 0; } $ g++ test.cpp; ./a.out e10 = \clifford[0]{e}^{{0} } e20 = \clifford[0]{e}^{{1} } e10 = e~0 e20 = e~1

How can I deal with this?.

May be make both formats print the same (well, almost) thing, like this patch does: diff --git a/ginac/clifford.cpp b/ginac/clifford.cpp index c7ce8d3..56ae54f 100644 --- a/ginac/clifford.cpp +++ b/ginac/clifford.cpp @@ -203,6 +203,7 @@ void clifford::do_print_dflt(const print_dflt & c, unsigned level) const seq[0].print(c, precedence()); c.s << "\\"; } else + c.s << "e[" << int(representation_label) << "]"; this->print_dispatch<inherited>(c, level); } Best regards, Alexei -- All science is either physics or stamp collecting.