> but it does guarantee that it could be. in which sense do you mean this? consider for instance constexpr int foo(int x) { if(x < 0) throw "error";

> but it does guarantee that it could be. in which sense do you mean this? consider for instance constexpr int foo(int x) { if(x < 0) throw "error"; return x; } I would say that "guarantee that it could be" means that as long as the function definition itself "builds" / is validated by the compiler, you can use it in a constexpr context yet this is not the case here: constexpr int a = foo(123); // works fine constexpr int b = foo(-123); // compile error int c = foo(-123); // works fine so having "constexpr" in the API does not mean that your code will always build ; as soon as you have constexpr the entire implementation of the function is part of the API, thus making the keyword moot like gpderetta says