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() and Assert() functions. To organize things nicely, I enclose all the debugging stuff in a Debug namespace.

It would make more sense for the Assert() function to also provide a source file and line number, which is only possible using the __LINE__ and __FILE__ macros. However, it is pretty unpleasant, inefficient etc... to always have to specify these two parameters.

So this is how my code would look like:

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

My question is, is it possible to place a macro which includes those two parameters inside the Debug namespace? Like this:

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");

// Output: Assertion failed on line 10 in file test.cpp:
//           Some_condition should be true

Is this valid c++? If not, is there any way of making this work?

Answer 1

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

No.

#define is a preprocessor directive. The macros are being replaced before anything else apart from removing comments (which means, before compilation). So at the time macros are replaced, the compiler knows nothing about your namespaces.

As other people state, in your case it will be fine. However, This is how you can get problems:

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

}
namespace B
{
 void Assert_ (int condition)
 {
     std::cout << "B";
 }
   #define Assert(a,b) Assert_(a)

}

int main(int argc, char *argv[])
{
    A::Assert(0,"asdasd");
    B::Assert(0,"asdasd");
}

So while it looks like the defines are "in the namespaces", they are not, and the last #define will be always be used, which in this case will lead to a compile-time error, because the code in main will be replaced by:

A::Assert(0);
B::Assert(0);

instead of

A::Assert(0,"asdasd", _FILE_, _LINE_);
B::Assert(0);

Answer 2

No, the preprocessor doesn't care about namespaces at all. In fact, the preprocessor runs, at least conceptually, before the compiler sees anything.

For myself, I just do a standard ASSERT macro, and expect that no "sane namespace" has something called ASSERT. Problem solved. Should I require a library that has an ASSERT of its own then I can still decide how to deal with this; however, the only library that I'm currently using with its own "assert" calls it BOOST_ASSERT or something like that...

Answer 3

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).

Answer 4

You can try __PRETTY_FUNCTION __ macro to print all the namespaces including function arguments.

Answer 5

Yes, and your macro would expand to exactly what you expect.

Debug::Assert (some_condition, "Some_condition should be true");

would be replaced by

Debug::Assert_(some_condition, "Some_condition should be true", __FILE__, __LINE__)