Question

The title is pretty self-explanatory, but here’s a simplified example:

#include <cstdio>

template <typename T>
struct MyTemplate {

    T member;

    void printMemberSize() {
        printf("%i\n", sizeof(T));
    }

};

int main() {

    MyTemplate<struct { int a; int b; }> t; // <-- compiler doesn't like this

    t.printMemberSize();

    return 0;

}

The compiler complains when I try to use an anonymous struct as a template argument. What’s the best way to achieve something like this without having to have a separate, named struct definition?

Answer 1

You are not allowed to define an unnamed type as a template argument in C++03 or even in C++0x.

The best you can do it to create a named struct local to main (in C++0x¹)

_{1 : You are not allowed to use a local type as template argument in C++03, however C++0x allows it.}

Also check out the Defect Report here. The proposed solution mentions

The following types shall not be used as a template-argument for a template type-parameter:

• a type whose name has no linkage
• an unnamed class or enumeration type that has no name for linkage purposes (7.1.3 [dcl.typedef])
• a cv-qualified version of one of the types in this list
• a type created by application of declarator operators to one of the types in this list
• a function type that uses one of the types in this list

---

The compiler complains when I try to use an anonymous struct as a template parameter.

Did you mean template argument? Template parameter is different from template argument.

For example

template < typename T > // T is template parameter
class demo {};

int main()
{
   demo <int> x; // int is template argument
}