On Thu, 2007-05-24 at 14:50 -0400, Warren Weckesser wrote:
This works for me:
string s = ex_to<symbol>(E).get_name();
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 wwhen I do cout << ex_to<symbol>(e00).get_name() << endl; Anything gets plotted. So I suspect that e00 is not a symbol an then, that I can not change its name.??? It woul be interesting to have this posibility, tough elements of diferent basis are right now plotted in the same way. 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. How can I deal with this?. Thanks in advance Javier