Why Does Partial Specialization of a Template Member Function Fail with an Incomplete Type Error?

Errors Related to Incomplete Types and Partial Template Specialization

When attempting to partially specialize the member function bar of a template class foo, users may encounter an error indicating the invalid use of an incomplete type. In particular, the following code snippet generates an error:

template <typename s typename t>
struct foo {
   void bar();
};

template 
void foo<int t>::bar() {
}</int></typename>

This is because the compiler cannot determine the complete type of the partially specialized template class before encountering the specialized function definition. To resolve this issue, one must partially specialize the entire template class, as shown below:

template <typename s typename t>
struct foo {
   void bar() {
       // If S is not int, do default behavior for non-int S
       if constexpr(!std::is_same<s int>::value) {
          // Default behavior
       }
       // If S is int, do specialized behavior unique to int
       else {
          // Int-specific behavior
       }
   }
};</s></typename>

By partially specializing the entire template class, the compiler can infer the complete type before encountering the specific function definition. This allows the specialized function to access and modify members specific to the newly introduced specialization.

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