This rule is part of MISRA C++:2023.
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Rule 17.8.1 - Function templates shall not be explicitly specialized
Category: Required
Analysis: Decidable,Single Translation Unit
Amplification
This rule also applies to member function templates, but not non-template member functions of class templates.
Rationale
Explicit function specializations will be considered only after overload resolution has chosen a best match from the set of primary function
templates. Furthermore, when the overload set contains both template and non-template versions that are otherwise an equal match for overload
resolution, the template version (and therefore its specializations) will not be selected. All of this may be inconsistent with developer
expectations.
Note: overloads provide a better solution than the use of explicit function specializations.
Example
template< typename T > void f1( T ); // Overload # 1A
template<> void f1< char * >( char * ); // Non-compliant - explicit
// specialization of overload # 1A
template< typename T > void f1( T * ); // Overload # 1B
template< typename T > void f2( T ); // Overload # 2A
template< typename T > void f2( T * ); // Overload # 2B
void f2( char * ); // Overload # 2C - rule does not apply
template< typename T > void f3( T );
template<> void f3< char * >( char * ); // Non-compliant - explicit
// specialization of f3
void b( char * s )
{
f1( s ); // Calls overload # 1B, with T = char
f2( s ); // Calls overload # 2C
}
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