Introduction / Context:
Templates enable writing type-independent code. When you instantiate a class template with different type arguments (e.g., MyBox and MyBox), you get distinct types with the same interface. Understanding how to invoke their member functions clarifies generic programming workflows in C++.
Given Data / Assumptions:
- We have a class template with the same set of member functions for each instantiation.
- Two instantiations are created: one with int and one with double.
- The question concerns the procedure to call a member function on each object.
Concept / Approach:
- Each instantiation is a separate type (e.g., Box vs Box), but the syntax to call members is identical: object.member(args).
- No language rule requires prefixing calls with the type name; overload resolution and templates handle this automatically.
- Availability of a function on one instantiation does not remove it from another; interfaces are replicated per instantiation, subject to constraints.
Step-by-Step Solution:
Create objects: Box bi; Box bd;.Invoke: bi.push(1); bd.push(1.0); The call syntax is the same.Therefore, there is no difference in the procedure to call a member function across instantiations.
Verification / Alternative check:
Examine the standard library: std::vector and std::vector use identical call syntax for size(), push_back(), etc.
Why Other Options Are Wrong:
- Prefix words int/double: Not part of call syntax.
- Becomes unavailable: False; instantiations coexist independently.
- Cannot perform: Templates are designed for exactly this usage.
Common Pitfalls:
- Assuming templates erase types at compile time like some languages; in C++, types remain distinct.
- Forgetting that constraints (e.g., operations required on T) may limit certain instantiations.
Final Answer:
There is no difference in the procedure to call a member function
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