在不需要时使用“template”和“typename”消歧器

Question

This question covers when and why the typename and template disambiguators are needed in C++ template code.

Is it valid to use these disambiguators in cases where they are not needed in C++03? How about in C++11?

Answer 1

It's valid in conforming C++03/C++11 compilers, for some definition of "valid."

C++03 ISO/IEC 14882:2003 §14.2.5:

[ Note: just as is the case with the typename prefix, the template prefix is allowed in cases where it is not strictly necessary; ie, when the expression on the left of the -> or . , or the nested-name-specifier is not dependent on a template-parameter . ]

C++11 ISO/IEC 14882:2011 §14.2.5:

[ Note: As is the case with the typename prefix, the template prefix is allowed in cases where it is not strictly necessary; ie, when the nested-name-specifier or the expression on the left of the -> or . is not dependent on a template-parameter , or the use does not appear in the scope of a template. — end note ]

Note that you can't use template when the member in question isn't actually a template—you aren't allow to lie with it. Also note that for typename , the type has to be a qualified type (eg X::Y , not just X ). C++11 also changed it so that you don't have to be in the scope of a template, whereas C++03 required you to be in a template. Also note that compilers are likely to differ on whether they actually let you do this. Under Clang, for instance, this warns under the flag -Wc++11-extensions .

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Here are some examples, assuming the following definition:

struct X {
    typedef int Y;
    template <typename T> static void foo();
    static void bar();
    template <typename T> static void baz(T);
};

Invalid in both C++03 and C++11:

template <typename T>
void foo() {
    typename int z = 0; // int is not a qualified name.
    X::template bar();  // X::bar is not a template.
    X::template baz(z); // no template argument list.
}

Invalid in C++03, valid in C++11 (but produces a warning on my copy of Clang):

void bar() {
    typename X::Y z = 0;    // not in the body of a template, so
    X::template foo<int>(); // no possibility of dependent names.
}

Valid in both C++03 and C++11:

template <typename T>
void baz() {
    typename X::Y z = 0;    // not a dependent name, so 'typename'
    X::template foo<int>(); // isn't strictly necessary.
}