为什么函数模板的显式实例化不使用内联或constexpr

Question

Referring to cppreference's section on function templates :

Explicit instantiation of a function template or of a member function of a class template cannot use inline or constexpr

These topics, inline and constexpr , seem separate and unrelated. Why does this restriction exist?

Answer 1

Because they serve opposite purposes.

The point of explicit instantiation is, in a source file, to provide definitions for templates that your project needs so that you don't have to fully define the template in your header file .

The point of inline is to allow function definitions in a header - so that multiple definitions of the function across multiple translation units can be collapsed into one.

constexpr functions must have definitions visible for the compiler to actually be able to invoke them at compile-time. There is no link-time constexpr .

It doesn't make sense to explicitly instantiate an inline or constexpr function - those function templates must already be defined in header files and so will be able to be implicitly instantiation on-demand.