#define CreateObjName(or,n,c) {(or) = ((n)##(c)); ++(c);}
fGameObject::fGameObject() : m_id(id_count), m_type(OBJ_DEFAULT_OBJ)
{
CreateObjName( m_name, default_name, obj_count );
++id_count;
}
// m_name is a const char*
// default_name is a const char*
// obj_count is an unsigned int
fGameObject::fGameObject() : m_id(id_count), m_type(OBJ_DEFAULT_OBJ){ {(m_name) = ((default_name)(obj_count)); ++(obj_count);} ++id_count;}Quote:Original post by ETFBladeI gather from the syntax that this is C++, in which case you should really be using std::string. With a little help from Boost, your example becomes:
I take it the ## operator is then the wrong thing to use in this situation?
m_name = default_name + boost::lexical_cast<std::string>(obj_count++);#define FUNCTION_TIMER_BEGIN() static int __FUNCTION__##_timer = 0; static int __FUNCTION__##_total_time = 0; timer = time(NULL);#define FUNCTION_TIMER_END() __FUNCTION__##_total_time += time(NULL) - __FUNCTION__##_timer; __FUNCTION__##_timer = 0;Quote:Original post by deathkrush
The ## operator is only useful for generating variable names, function names or other things that need to be done at compile time.
Quote:Original post by Zahlman
The *preprocessor* is the wrong thing to be trying to use, because it operates on the *text of your program*, *before compile time* and *has actually has almost no concept of the C++ language at all*.
Quote:Original post by Oluseyi
Once more, with feeling: the preprocessor operates on the text of your program before compile time.
int max(int x, int y){ return 0;}#define max(x,y) (((x) > (y)) ? (x) : (y))#include <iostream>int main(){ std::cout << (max)(4,5); std::cout << std::endl; std::cout << max(4,5);}
