void* l = dlsym(lib,"_ZN11Environment9LibLogger14log_processingEiNSt7__cxx1112basic_stringIcSt11char_traitsIcESaIcEEEjS6_z");
*(void **)&log_fcn = l;
std::cout<<"Address"<<<<l<<"LOG_FCN "<<log_fcn<<std::endl;
I am trying to print address of l
and log_fcn
but both are not same. Why is that, and how can I get address of dlsym
assign to function pointer?
Output example:
l= 0x7efe10eabaa0 LOG_FCN 1
void (*log_fcn)(int level, std::string frmt, unsigned int line_no, std::string file_name, ...); function decleration
CodePudding user response:
There is an operator <<
for void*
, but not for function pointers.
Function pointers are implicitly converted to bool
, not void*
, and bool
prints as 0
or 1
by default.
Your 1
output indicates that log_fcn
is not the null pointer.
Convert it when printing:
std::cout << "Address" << l << "LOG_FCN "<< reinterpret_cast<void*>(log_fcn) << std::endl;
I would recommend that you also do the assignment in the conventional form, by casting the right-hand side:
log_fcn = reinterpret_cast<decltype(log_fcn)>(l);