How Does Partial Ordering Determine Specialization Between Function Templates?

Partial Ordering Procedure in Template Deduction

The partial ordering procedure in template deduction determines the specialization relationship between two function templates. It involves two stages:

Stage 1: Transformation

For each template, the partial ordering procedure creates a "transformed function type" by replacing all type, non-type, and template template parameters with unique, unused types.

Stage 2: Comparison

The transformed function types are compared in two ways:

• Forward Matching: The transformed function type of template 1 is matched against the original function type of template 2.
• Backward Matching: The transformed function type of template 2 is matched against the original function type of template 1.

If one of the matches succeeds and the other fails, then the template with the successful match is considered more specialized. If neither match succeeds, then neither template is more specialized.

Example:

Consider the following two function templates:

template<typename t typename u>
void foo(T, U); // Template 1

template<typename t>
void foo(T const*, X<t>); // Template 2</t></typename></typename>

To determine which template is more specialized, we:

1. Create Transformed Function Types:

• Template 1 transformed: void foo(int, bool)
• Template 2 transformed: void foo(char const*, X)

2. Compare Transformed Function Types:

• Forward Matching: void foo(int, bool) cannot be matched to void foo(T const*, X).
• Backward Matching: void foo(char const*, X) can be matched to void foo(T, U) by deducing T = char const* and U = X.

Conclusion: Template 2 is more specialized than Template 1.

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