How to use token concatenation with scope operator? - c++

Question

I would like to do something like the following:
#define CODE_BLOCK( id) \
do { \
someNameSpace::##id foo; \
foo.ParseFromString( ...); \
print_##id( foo);
} while(0)
[...]
CODE_BLOCK( BlahStream);
So basically, I want foo to be declared as type someNameSpace::id where id macro argument. print_id is defined elsewhere. Anyways, g++ complains with the message:
error: pasting "::" and "BlahStream" does not give a valid preprocessing token
Is there anyway around this?
If it helps, foo is from some userdefined googleprotocol buffer. My g++ version is 4.7.2 on debian linux.
Thanks.

Answer 1

Just remove the ## operator and it'll work just fine.
e.g.
#define stuff(x) \
std:: x
#include <iostream>
int main() {
stuff(cout) << "Hello World\n";
}

Answer 2

The preprocessor is already aware that x::y is three separate tokens: x, :: and y.
Therefore, you do not need any concatenation and, in fact, as you've found out, trying to use it will result in an error.
When you write this:
someNameSpace::##id
you are writing someNamespace, ::y … and ::y is not a single token.
You should simply write:
someNameSpace::id
Now you have someNamespace, ::, y … as desired.
Here's your finished code:
#define CODE_BLOCK(id) \
do { \
someNameSpace::id foo; \
foo.ParseFromString(...); \
print_##id(foo);
} while(0)
// [...]
CODE_BLOCK(BlahStream);
N.B. I've removed the spaces that you keep writing after the ( symbol, because ew. :)