CODEWITHSUNDEEP
c++namespacestypedeflanguage-lawyername-lookup
Supremum
Supremum

Reputation: 552

Different behaviour between Clang and GCC when performing qualified name lookup

Consider the following program:

#include <iostream>

namespace N {
    int j = 1;
}

namespace M {
    typedef int N;
    void f() {
        std::cout << N::j << std::endl;
    }
}

int main() { M::f(); }

Compiling it with clang gives the following compiler error:

prog.cc:10:22: error: 'N' (aka 'int') is not a class, namespace, or
enumeration
    std::cout << N::j << std::endl;
                 ^ 1 error generated.

GCC does not give any compiler error. I'm trying to figure out for what compiler I should file the bug report for. Which compiler has the correct behaviour and why (references to the c++ standard)?

Wandbox - Clang: http://melpon.org/wandbox/permlink/s0hKOxCFPgq5aSmJ

Wandbox - GCC: http://melpon.org/wandbox/permlink/i2kOl3qTBVUcJVbZ

Upvotes: 9

Views: 456

Answers (1)

Angew is no longer proud of SO
Angew is no longer proud of SO

Reputation: 171127

Clang is correct on this one. Quoting C++11, 3.4.3/1 [basic.lookup.qual]:

... If a :: scope resolution operator in a nested-name-specifier is not preceded by a decltype-specifier, lookup of the name preceding that :: considers only namespaces, types, and templates whose specializations are types. If the name found does not designate a namespace or a class, enumeration, or dependent type, the program is ill-formed.

Per this clause, types are supposed to be considered during the lookup, so the typedef N should be found. And since it does not designate a namespace, class, enumeration, or dependent type, the program is ill-formed.

Upvotes: 11

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