How Can I Resolve \'Non-type Template Argument is Not a Constant Expression\' Errors in Constexpr Functions with References?

Referencing Non-Constant Parameters in Constexpr Functions

Consider the following function template:

template <size_t S1, size_t S2>
auto concatenate(const std::array<uint8_t, S1>& data1,
                 const std::array<uint8_t, S2>& data2)
{
    // Error in constexpr contexts due to reference parameter
    std::array<uint8_t, data1.size() + data2.size()> result;

    // ...
}

Using Clang 6.0 with C 17, this code fails to compile with the error "non-type template argument is not a constant expression." Evaluating references within constexpr functions is problematic due to potential aliasing issues.

Understanding the Standard

According to the C Standard ([expr.const]/4), an expression cannot be a core constant expression if it evaluates to a reference that:

• Does not have a preceding initialization
• Was created before the evaluation of the constant expression

Solution

To resolve this issue, one can substitute the problematic reference parameter with its corresponding template parameter:

template <size_t S1, size_t S2>
auto concatenate(const std::array<uint8_t, S1>& data1,
                 const std::array<uint8_t, S2>& data2)
{
    // Problem solved by using S1 + S2 instead
    std::array<uint8_t, S1 + S2> result;

    // ...
}

By using the template parameters directly, we eliminate the need for evaluating references in constexpr contexts and ensure that the function can be used in constant expression evaluations.

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