发生此错误是因为您的朋友声明充当另一个未模板化的 f 函数的函数声明.

The error occurs because your friend declaration acts as a function declaration of another non-templated f function.

您必须像这样声明它，以便告诉编译器它是模板函数:

You have to declare it like this in order to tell the compiler that it is a template function:

    friend void f<T>(T);

请考虑以下示例:

template<class T>
struct S {
    friend void foo(int);
};

int main() {
    S<int> s;
    foo(42);
}

这将引发链接器错误，提示此处未解析的外部符号 foo .在这种情况下，声明了 foo ，但未通过好友声明进行定义.

This will throw a linker error muttering about an unresolved external symbol foo here. In this case foo is declared but not defined through the friend declaration.

如果我们现在注释掉 S< int>s; 我们现在不是链接器，而是编译器错误:'foo':找不到标识符，因为尚未将 foo 声明为S< int> 未编译.

If we now comment out S<int> s; we now get not a linker but a compiler error: 'foo': identifier not found, because foo has not been declared, as S<int> isnt compiled.