Why string concat macro doesn't work for this "+" case? - c++

Question

Short question:
Is it permitted to concat special signs such as +, - for the string concatenation macro ##? For example,
#define OP(var) operator##var
will OP(+) be expanded to operator+?
Exact problem:
#include "z3++.h"
#include <unordered_map>
namespace z3 {
z3::expr operator+(z3::expr const &, z3::expr const &);
}
typedef z3::expr (*MyOperatorTy)(z3::expr const &, z3::expr const &);
#define STR(var) #var
#define z3Op(var) static_cast<MyOperatorTy>(&z3::operator##var)
#define StrOpPair(var) \
{ STR(var), z3Op(var) }
void test() {
std::unordered_map<std::string, MyOperatorTy> strOpMap1{
{"+", static_cast<MyOperatorTy>(&z3::operator+)}}; // fine
std::unordered_map<std::string, MyOperatorTy> strOpMap2{StrOpPair(+)}; // error
}
For strOpMap2, using clang++ -c -std=c++11, it reports:
error: pasting formed 'operator+', an invalid preprocessing token
while using g++ -c -std=c++11, it gives:
error: pasting "operator" and "+" does not give a valid preprocessing token
By reading the manual by gcc I find it should be possible to concat, but why both compilers emit errors?

Answer 1

You can paste punctuation to form other punctuation, e.g.
#define PASTE(a,b) a##b
int main()
{
int i = 0;
i PASTE(+,+);
// i == 1 now
}
The ## operator is for producing a valid preprocessing token from other preprocessing tokens. The result of pasting must be a valid preprocessing token. So this is not valid:
PASTE(i,++)
because i++ is not a preprocessing token; it's two adjacent tokens i and ++.
The list of possible tokens is (N3797):
header-name
identifier
pp-number
character-literal
user-defined-character-literal
string-literal
user-defined-string-literal
preprocessing-op-or-punc
each non-white-space character that cannot be one of the above
Note: at the preprocessing stage, keyword does not exist; but after preprocessing, any identifiers which should be keyword are converted (semantically) into keywords. So you can build keywords by pasting shorter words.
In your code, operator+ is two tokens: operator and +. So you do not build it with ##; you just do one then the other.
#define OP(punc) operator punc