constexpr function parameters as template arguments

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You tell the compiler, that addFunc would be a constexpr. But it depents on parameters, that are not constexpr itself, so the compiler already chokes on that. Marking them const only means you are not going to modify them in the function body, and the specific calls you make to the function are not considered at this point.

There is a way you can make the compiler understand you are only going to pass compile time constants to addFunc: Make the parameters a template parameters itself:

template <int x, int y>
constexpr int addFunc() {
    return add<x,y>::ret;
}

Then call as

cout << addFunc<1,2>() << endl;

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