Why is a constexpr function on a reference not constexpr?

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Because you have evaluated a reference. From [expr.const]/4:

An expression e is a core constant expression unless the evaluation of e, following the rules of the abstract machine, would evaluate one of the following expressions:

  • …
  • an id-expression that refers to a variable or data member of reference type unless the reference has a preceding initialization and either
    • it is usable in constant expressions or
    • its lifetime began within the evaluation of e;
  • …

Your reference parameter has no preceding initialization, so it cannot be used in a constant expression.

You can simply use S1 + S2 instead here.

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