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Why Does Template Type Deduction Fail When Inferring from a Member Type?

Barbara Streisand
Barbara StreisandOriginal
2024-12-19 09:24:10136browse

Why Does Template Type Deduction Fail When Inferring from a Member Type?

Deduction Failure in Template Type Inference

When attempting to deduce template arguments from a type member, it's crucial to ensure the context is deducible. In the given code, the error occurs within the test function when the compiler tries to deduce the template argument T from typename TMap::Type.

This failure stems from the concept of "non-deducible context." In this case, the template type TMap's member type 'Type' has no inherent relationship with the template argument T.

Consider a hypothetical specialization of TMap:

template <>
struct TMap<SomeType> {
    typedef std::map<double, double> Type;
};

If the compiler had to deduce T from TMap::Type, it would encounter an ambiguity. The value of TMap::Type is std::map. However, it's not guaranteed that T in TMap is the same as SomeType in the specialization.

Further specializing TMap:

template <>
struct TMap<OtherType> {
    typedef std::map<double, double> Type;
};

The situation worsens, as now the following holds:

  • TMap::Type = std::map
  • TMap::Type = std::map

Given this ambiguity, the compiler cannot determine whether T is SomeType or OtherType. It also cannot determine the number of possible choices or identify those choices.

Therefore, in contexts like this where the template member type has no direct relationship with the template argument, template argument deduction fails, and the compiler raises an error.

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