Hi, all. It seems that numeric conversion routines like to_long()
don't check for imaginary numbers, and happily return random
results. Here's an example program:
#include <ginac/ginac.h>
#include <ginac/parser.h>
#include <fstream>
#include <sstream>
using namespace std;
using namespace GiNaC;
int
main()
{
parser reader;
istringstream ifs("I");
ex expr = reader(ifs);
if (is_a<numeric>(expr)) {
numeric x = ex_to<numeric>(expr);
cout << "Numeric: " << x << "\n";
cout << ".real(): " << x.real() << "\n";
cout << ".imag(): " << x.imag() << "\n";
cout << ".numer(): " << x.numer() << "\n";
cout << ".denom(): " << x.denom() << "\n";
cout << ".numer().to_long(): " << x.numer().to_long() << "\n";
cout << ".denom().to_long(): " << x.denom().to_long() << "\n";
cout << ".to_long(): " << x.to_long() << "\n";
} else {
cout << "Not numeric: " << expr << "\n";
}
}
The result of it with GiNaC 1.7.8 (CLN 1.3.6) is:
$ c++ -o example example.cpp -lginac -lcln && ./example
Numeric: I
.real(): 0
.imag(): 1
.numer(): I
.denom(): 1
.numer().to_long(): 9
.denom().to_long(): 1
.to_long(): 9
Seeing that the documentation contains this phrase:
Both to_int()/to_long() and to_double() discard the imaginary
part of complex numbers.
... I'm guessing this is not the expected behavior.
* * *
As a related question, is there any way to prevent parser from
interpreting "I" as an imaginary unit? I'd like it to be just a variable.