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

Short answer: there are no constexpr function parameters in C++11/14.

Longer answer: in test1(), if i is not a compile-time constant, the function is still usable at run-time. But in test2(), it cannot be known to the compiler whether i is a compile-time constant, and yet it is required for the function to compile.

E.g. the following code for test1 will compile

int i = 0;    
char a = test1("Test", i); // OK, runtime invocation of test1()

constexpr int i = 0;
constexpr char a = test1("Test", i); // also OK, compile time invocation of test1()

Let's simply your test2() to

constexpr char test3(unsigned i)
{
    return t::value;
}

This will not compile for test3(0) because inside test3(), it cannot be proven that i is an unconditional compile-time expression. You would need constexpr function parameters to be able to express that.

Quote from the Standard

5.19 Constant expressions [expr.const]

2 A conditional-expression e is a core constant expression unless the evaluation of e, following the rules of the abstract machine (1.9), would evaluate one of the following expressions:

— an id-expression that refers to a variable or data member of reference type unless the reference has a preceding initialization and either
— it is initialized with a constant expression or

— it is a non-static data member of an object whose lifetime began within the evaluation of e;

This section has the following code example corresponding to your question:

constexpr int f1(int k) {
    constexpr int x = k; // error: x is not initialized by a
                         // constant expression because lifetime of k
                         // began outside the initializer of x
    return x;
}

Because x in the above example is not a constant expression, it means that you can't instantiate templates with either x or k inside f1.