问题
In trying to compile this program:
namespace MyNamespace {
template<typename T>
class Test {
public:
class Inner {
int x;
public:
Inner() : x(0) { }
friend Inner& operator++(Inner& rhs);
};
Inner i;
};
}
template<typename T>
typename MyNamespace::Test<T>::Inner& operator++(typename MyNamespace::Test<T>::Inner& rhs) {
rhs = MyNamespace::Test<T>::Inner(rhs.x + 1);
return rhs;
}
int main() {
MyNamespace::Test<int> t;
MyNamespace::Test<int>::Inner i = t.i;
++i;
}
I get the error
unresolved external symbol "class MyNamespace::Test::Inner & __cdecl MyNamespace::operator++(class MyNamespace::Test::Inner &)" (??EMyNamespace@@YAAAVInner@?$Test@H@0@AAV120@@Z) referenced in function _main
Which is weird because that's the exact signature of the non-member friend function operator++
that I defined. How do I fix this? And I do not have the option of including in as a member function because I need to change the object that the operand is referring to without using a copy constructor (because there is no copy constructor).
Update:
If I add template<typename T>
above the friend Inner&...
, I get the errors
could not deduce template argument for 'T' 1>
main.cpp(21) : see declaration of 'operator
++'
error C2783:
'MyNamespace::Test<T>::Inner &MyNamespace::operator++(MyNamespace::Test<T>::Inner &)' : could not deduce template
argument for 'T' with
[
T=int
]
main.cpp(13) : see declaration of
'MyNamespace::operator ++'
main.cpp(30): error C2675: unary '++' : 'MyNamespace::Test<T>::Inner' does not define this operator or a
conversion to a type acceptable to the predefined operator
with
[
T=int
]
回答1:
Why do you think it can't be a member function? An instance member should work just fine:
namespace MyNamespace
{
template<typename T>
class Test
{
public:
class Inner
{
int x;
public:
Inner& operator++( void ) { ++x; return *this; }
};
Inner i;
};
}
This doesn't require a copy constructor.
Defining a friend should work too, as long as the definition is with the friend declaration:
namespace MyNamespace
{
template<typename T>
class Test
{
public:
class Inner
{
int x;
public:
friend Inner& operator++( Inner& operand ) { ++operand.x; return operand; }
};
Inner i;
};
}
The friend function will be placed at namespace scope, according to [class.friend]
来源:https://stackoverflow.com/questions/6974319/unresolved-external-symbol-with-operator-overloading-and-templates