Consider these two:
namespace X1 { A operator "" _x(unsigned long long i) { return A{i}; } }; namespace X2 { B operator "" _x(unsigned long long i) { return B{i}; } }; The x literal is defined twice, but one of them is defined in namespace X1 while another is defined in namespace X2.
According to the C++ standard, can this code be compiled?
A user-defined literal is treated as a call to operator ""X(...) where X is an identifier (for example, _x) and the ... depends on the form of the user-defined literal. Lookup of the appropriate user-defined literal operator then proceeds according to the usual rules for unqualified lookup; [lex.ext]:
2 - A user-defined-literal is treated as a call to a literal operator or literal operator template (13.5.8). To determine the form of this call for a given user-defined-literal
Lwith ud-suffixX, the literal-operator-id whose literal suffix identifier isXis looked up in the context ofLusing the rules for unqualified name lookup (3.4.1). [...]
The definitions of the _x literal only conflict if both definitions are made available for unqualified lookup by a using or using namespace declaration or by entering either namespace; the conflict is only problematic if a call to the unqualified operator is actually made.
{ using namespace X1; auto o = 5_x; } // OK; o is of type A { using namespace X2; auto o = 5_x; } // OK; o is of type B { using namespace X1; using namespace X2; } // OK; operator "" _x is not used { using namespace X1; using namespace X2; auto o = 5_x; } // ambiguous It would also be OK if the _x were different types of user-defined literal operators e.g. if one were an integer literal operator and the other a float literal operator.
using usage would be using X1::operator"" _x; and using X2::operator"" _x; in case anyone finds this question and is wondering.
unsigned long longparameter rather thanint.