I am asking myself why the following code works and what the specifier extern
does when instantiating baz_instance
:
struct baz {
int value;
};
extern const baz baz_instance = {3};
template<baz const& b>
int foo(){
return b.value;
}
int main(){
foo<baz_instance>();
return 1;
}
Why does the above code compile in the first place and why doesn't it compile anymore if the extern
specifier is left out? What does the extern
specifier do in this example?
The extern
keyword means that it will have external linkage, in other words the symbol will be exported when the translation unit will be compiled. Because your type is const, it by default has internal linkage (as though it has been declared static
). Templates cannot depend on types that only have internal linkage.
I would love to know the reason why, but it seems like it's lost to the sands of time: Why did C++03 require template parameters to have external linkage?.
This is one of the parts of the standard that changed from C++03 to C++11.
In C++03, [temp.arg.nontype] reads:
A template-argument for a non-type, non-template template-parameter shall be one of:
- [...]
- [...]
- the address of an object or function with external linkage, including function templates and function template-ids but excluding non-static class members, expressed as & id-expression where the & is optional if the name refers to a function or array, or if the corresponding template-parameter is a reference; or
- [...]
In C++11, that got updated as a result of issue 1155, though GCC still has a bug with regards to this behavior:
- a constant expression (5.19) that designates the address of a complete object with static storage duration and external or internal linkage or a function with external or internal linkage, including function templates and function template-ids but excluding non-static class members, expressed (ignoring parentheses) as & id-expression, where the id-expression is the name of an object or function, except that the & may be omitted if the name refers to a function or array and shall be omitted if the corresponding template-parameter is a reference; or
In C++14, that got simplified even further and doesn't even mention linkage.
As to your specific question, the extern
specifier adds external linkage to baz_instance
. Without it, baz_instance
has internal linkage. In C++03, you needed external linkage to have a non-type template parameter of reference type. In C++11, you don't anymore - so extern
is no longer necessary and it compiles fine without it.
From 14.3.2.1 the standard states:
A template-argument for a non-type, non-template template-parameter shall be one of:
- the address of an object or function with external linkage, including function templates and function template-ids but excluding non-static class members, expressed as & id-expression where the & is optional if the name refers to a function or array, or if the corresponding template-parameter is a reference;
From https://stackoverflow.com/a/643927/451600 by mweerden
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