Why is this not a constant expression?

Question

In this trivial example, test2 fails to compile even though test1 succeeds, and I don\'t see why that is the case. If arr[i] is suitab

Answer 1

There's a misconception of what constexpr does here. It indicates that a function must be evaluatable at compile time for suitable arguments, but it does not remove the requirement still to compile in the general case.

Let's take the first version:

template 
constexpr char test1(const char (&arr)[N], unsigned i) {
    return arr[i];
}

Now, this is clearly a compile-time evaluation:

enum { CompileTimeConstant = test1("Test", 0) };

your example may be, but it's an optimizer/QoI issue:

char MayBeCompileTimeConstant = test1("Test", 0);

and this example obviously isn't, but is still required to be evaluatable

char arr[10];
int i;
std::cin >> i;
std::cin >> arr;
char b = test1(arr, i);
std::cout << "'" << arr << "'[" << i << "] = " << b << '\n';

Since test2 can't possibly compile in for the last case, it can't compile at all. (Please note I'm not suggesting that code is good).