constexpr function parameters as template arguments

Question

I am playing around with some toy code using c++11 to figure out a bit more about how things work. During this I came across the following issue that simplifies down to:

template <int x, int y> class add { public:     static constexpr int ret = x + y; };  constexpr int addFunc(const int x, const int y) {     return add<x,y>::ret; }  int main() {     const int x = 1;     const int y = 2;     cout << add<x,y>::ret << endl; // Works     cout << addFunc(1,2) << endl;  // Compiler error     return 0; } 

I'm using GCC 4.8.1 and the output is:
'x' is not a constant expression in template argument for type 'int'
'y' is not a constant expression in template argument for type 'int'

What exactly is the difference between the two ways I am trying to calculate add::ret? Both of these values should be available at compile time.

Answer 1

If your purpose is just to shorten code a bit, in C++14 you can create variable template:

template <int x, int y> constexpr int addVar = x + y;  cout << addVar<5, 6> << endl; // Works with clang 3.5, fails on GCC 4.9.1 

GCC 5 will also support this.

Answer 2

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;