C++ templates specialization syntax

Solution 1:

Here are comments with each syntax:

void foo(int param); //not a specialization, it is an overload

void foo<int>(int param); //ill-formed

//this form always works
template <> void foo<int>(int param); //explicit specialization

//same as above, but works only if template argument deduction is possible!
template <> void foo(int param); //explicit specialization

//same as above, but works only if template argument deduction is possible!
template void foo(int param); //explicit instantiation

Added by me:

//Notice <int>. This form always works!
template void foo<int>(int param); //explicit instantiation

//Notice <>. works only if template argument deduction is possible!
template void foo<>(int param); //explicit instantiation

From coding point of view, overload is preferred over function-template-specialization.

So, don't specialize function template:

• Why Not Specialize Function Templates?
• Template Specialization and Overloading

And to know the terminologies:

• instantiation
• explicit instantiation
• specialization
• explicit specialization

See this :

• Difference between instantiation and specialization in c++ templates

Solution 2:

Using Visual Studio 2012, it seems to work slightly different if there's no function argument:

template <typename T> T bar( );
//template int bar<int>( ) { return 0; } doesn't work
template < > int bar<int>( ) { return 0; } //does work