Question

When compiling the code below with gcc, I get an error: ‘i’ cannot appear in a constant-expression.

Why is this?

#include <iostream>

using namespace std;
template<int p>
class C
{
public:
    void test();
};

template<int p>
void C<p>::test()
{
    p = 0;
}

char const *const p = "hello";
int main()
{
    const int i = (int)p;
    C<i> c;
}

Answer 1

The variable i is not mutable at runtime because it is const, but it is not a “constant expression” because it is not evaluated at compile-time.

(int)p; is a reinterpret_cast. Integral constant expressions cannot have reinterpret_cast. It is explicitly forbidden (§5.19/2):

A conditional-expression is a core constant expression unless it
involves one of the following as a potentially evaluated subexpression
(§3.2), but subexpressions of logical AND (§5.14), logical OR (§5.15),
and conditional (§5.16) operations that are not evaluated are not
considered [Note: An overloaded operator invokes a function.—end
note ]:

— […]

— a reinterpret_cast (§5.2.10);

— […]