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“template<>” vs “template” without brackets – what's the difference?

Posted by: admin November 30, 2017 Leave a comment


Suppose I’ve declared:

template <typename T> void foo(T& t);

Now, what is the difference between

template <> void foo<int>(int& t);


template void foo<int>(int& t);

semantically? And do template-with-no-brackets and template-with-empty-brackets have other semantics in other contexts?

Related to:
How do I force a particular instance of a C++ template to instantiate?


template <> void foo<int>(int& t); declares a specialization of the template, with potentially different body.

template void foo<int>(int& t); causes an explicit instantiation of the template, but doesn’t introduce a specialization. It just forces the instantiation of the template for a specific type.


With class/struct,

template <typename T> struct foo {};

Following is a specialization:

template <> struct foo<int>{};

Following is an explicit instantiation:

template struct foo<int>;