dear all,
is it possible to load a shared library made of some functions (no
classes) without making a dictionnary ? I would like to obtain the same
behavior than using .L source.cc but using .L libsource.so.
for instance:
cat test.cc:
#include <iostream>
void truc()
{
cout << "hello" << endl;
}
then build libtruc.so with
g++ test.cc -c -o test.o
then
g++ -shared test.o -o libtruc.so
then, in ROOT,
.L libtruc.so // is ok
but
truc() fails :
root [0] .L libtruc.so
root [1] truc()
Error: Function truc() is not defined in current scope
FILE:/tmp/fileYEaRUQ_cint LINE:1
Possible candidates are...
filename line:size busy function type and name
*** Interpreter error recovered ***
but
root [0] .L test.cc
root [1] truc()
hello
works fine !
Benoît
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