Given:
template<typename T>
void f( T ) {
}
enum { // if changed to "enum E" it compiles
e
};
int main() {
f( e ); // line 10
}
I get:
foo.cpp: In function ‘int main()’:
foo.cpp:10: error: no matching function for call to ‘f(<anonymous enum>)’
Yet if the enum
declaration is given a name, it compiles. Why doesn't it work for an anonymous enum? Ideally, I'd like it to promote the enum value e
to an int
an开发者_运维百科d instantiate f(int)
.
Unnamed type simply cannot be used as a template argument
C++03 says in 14.3.1[temp.arg.type]/2
A local type, a type with no linkage, an unnamed type or a type compounded from any of these types shall not be used as a template-argument for a template type-parameter.
This limitation was lifted in C++0x, and your program compiles with no diagnostics with MSVC++ 2010 and gcc 4.5.2 in C++0x mode.
Ideally, I'd like it to promote the enum value e to an int and instantiate f(int).
f(+e);
You can always explicitly do the promotion to clearly show your intention:
f(static_cast<int>(e));
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