CODEWITHSUNDEEP
c++namespacesname-lookup
TommiT
TommiT

Reputation: 609

Operator in namespace scope hiding another in global scope

Is this a compiler-bug?

template <typename T>
T& operator++(T& t)
{
    return t;
}

namespace asdf {

enum Foo { };
enum Bar { };

Foo& operator++(Foo& foo);

void fun()
{
    Bar bar;
    ++bar;
}

} // end namespace asdf

int main()
{
    return 0;
}

The GCC 4.7 error message is:

error: no match for 'operator++' in '++bar'
note: candidate is:
note: asdf::Foo& asdf::operator++(asdf::Foo&)
note: no known conversion for argument 1 from 'asdf::Bar' to 'asdf::Foo&'

It compiles if you comment out the line:

Foo& operator++(Foo& foo);

Upvotes: 13

Views: 886

Answers (3)

Bart van Ingen Schenau
Bart van Ingen Schenau

Reputation: 15758

No, this is not a compiler bug.

There are two name-lookups that get performed for the expression ++bar.

  • The regular name lookup searches the enclosing scopes and namespaces until it finds the first occurence of operator++. This search works inside out, so the global namespace is searched last. When looking for operator functions, member-functions are treated separately (and don't stop this search).
  • The argument-dependent lookup kicks in next and searches additional classes and namespaces, but only those that are related to the arguments of the function (operator++ in this case).

In the example in the question, the normal lookup finds asdf::operator++ and stops looking.
The argument-dependent lookup only adds the asdf namespace to the places to search, because that is the associated namespace for enum Bar. For that reason, the global operator++ can not be found.

You can make the global operator++ be found with a using declaration in namespace asdf.

Upvotes: 8

Pete Becker
Pete Becker

Reputation: 76245

Overloading only applies to names defined in the same scope. Once the compiler finds a matching name it doesn't look in outer scopes, even if the name it found applies to something that can't be used. This has nothing to do with operators; if the code used a function name in the same way that it uses operator++ it would get the same error. For example:

void f(int);

struct C {
void f(const C&);
void g() {
    f(3); // error: f(const C&) can't be called with argument 3
};

Upvotes: 1

Johannes Schaub - litb
Johannes Schaub - litb

Reputation: 506847

No that is not a bug. There are three parallel sets of operators considered. Members, non-member operators, and builtins.

The non-member ones are looked up by normal unqualified+ADL lookup, ignoring all class member functions. Hence the global operator is hidden by a lexical more closer one (and an intervening member function wouldn't have hidden other non-members).

Note that overload resolution takes place after name lookup1; in your case the name operator++ was found, but no appropriate overload.

If Bar had been declared globally, and/or the other operator in namespace asdf, ADL (in the former case) or ordinary unqualified lookup (in the latter case) would have dragged the operator in.

1: Overload resolution (...) takes place after name lookup has succeeded. (C++ Standard)

Upvotes: 14

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