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Is there any way to get behavior like this? // Some definition(s) of operator "" _my_str // Some definition of function or macro MY_STR_LEN using T1 = MY_STR_LEN("ape"_my_str); // T1 is std::integral_constant<std::size_t, 3U>. using T2 = MY_STR_LEN("aardvark"_my_str); // T2 is std::integral_constant<std::size_t, 8U>. It seems not, since the string literals are passed immediately to some_return_type operator "" _my_str(const char*, std::size_t); and never to a literal operator template (2.14.8/5). That size function parameter can't be used as a template argument, even though it will almost always be a constant expression. But it seems like there ought to be some way to do this. Update: The accepted answer, that this is not possible without an extra definition per literal, is accurate for C++11 as asked, and also C++14 and C++17. C++20 allows the exact result asked for: #include <cstdlib> #include <type_traits> #include <string_view> struct cexpr_str { const char* ptr; std::size_t len; template <std::size_t Len> constexpr cexpr_str(const char (&str)[Len]) noexcept : ptr(str), len(Len) {} }; // Essentially the same as // std::literals::string_view_literals::operator""sv : template <cexpr_str Str> constexpr std::string_view operator "" _my_str () noexcept { return std::string_view(Str.ptr, Str.len); } #define MY_STR_LEN(sv) \ std::integral_constant<std::size_t, (sv).size()>

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