Question

I have a template:

template<unsigned int N> struct IntN;

template <> struct IntN< 8> {
   typedef uint8_t type;
}; 

template <> struct IntN<16> {
   typedef uint16_t type;
};

And in main I initialise and alternate by doing this:

IntN< 8>::type c;

This seems to work, however, when I store the value inside a variable, it does not and I get the following error:

error: non-type template argument of type ‘int’ is not an integral constant expression

Here is an example of the code:

template<unsigned int N> struct IntN;

template <> struct IntN< 8> {
  typedef uint8_t type;
};

template <> struct IntN<16> {
   typedef uint16_t type;
};

int main(int argc, char *argv[]) {
int foo = 8;

IntN<foo>::type c;
}

Does anyone have any ideas? Thanks

Answer 1

The template arguments for integral template parameters must be constant expressions. An integral literal is a constant expression

IntN<8>::type c;

A constant variable initialized with a constant expression is a constant expression

const int n = 8;
IntN<n>::type c;

Here n is OK because it’s both const and initialized by a constant expression (8). The following will not compile though:

int n = 8;
const int m = n;
IntN<n>::type c; //error n is not const therefore not a constant expression
IntN<m>::type c; //error m is const but initialized with a non-constant expression therefore not a constant expression