Is it possible to place a macro in a namespace in c++?

Question

My application uses another output than the standard output for logging information, which is why I wrote my own Log(), Error(), Panic()

Answer 1

namespace Debug
{
    void Assert_(int condition, std::string message, std::string file, int line);
    #define Assert(a,b) Assert_(a, b, __FILE__, __LINE__)
}

// .... Somewhere where I call the function ....
Debug::Assert (some_condition, "Some_condition should be true"); 

This specific usage would do exactly what you want, but the Assert macro is in no way part of the Debug namespace... it's exactly as if you'd done:

namespace Debug
{
    void Assert_(int condition, std::string message, std::string file, int line);
}

#define Assert(a,b) Assert_(a, b, __FILE__, __LINE__)

// .... Somewhere where I call the function ....
Debug::Assert (some_condition, "Some_condition should be true"); 

Here, the substitution works not because Assert was in the Debug namespace (it's not in your code or this code, and the preprocessor has no clue what namespaces are about) - it works because Assert is recognised as an identifier for a macro, the substitution of Assert_ is made, then later the compiler proper happens to find there's a Debug::Assert_ So, say you have somewhere later in your translation unit you have some completely unrelated code:

my_object.Assert(my_functor);

The macro substituion will still kick in to produce a compile-time error saying you have the wrong number of arguments to a macro. Say the unrelated code was instead:

my_object.Assert(my_functor, "some text");

Then that would be replaced with:

my_object.Assert_(my_functor, "some text", __FILE__, __LINE__);

(Separately, it's standard practice not to use lower case letters in preprocessor macro names).