Why Doesn\'t `data1.size()` Work in a `constexpr` Function with Reference Parameters?

Constant Expressions in Constexpr Functions with Reference Parameters

Consider the following code snippet:

template <size_t s1 size_t s2>
auto concatenate(const std::array<uint8_t s1> &data1,
                 const std::array<uint8_t s2> &data2)
{
    std::array<uint8_t data1.size data2.size> result; // Error: non-type template argument is not a constant expression
    ...
}</uint8_t></uint8_t></uint8_t></size_t>

When compiled using clang 6.0 with -std=c 17, the function fails to compile due to the size member function of the array not being constexpr when applied to a reference.

Standard Rationale

The reason for this behavior is explained in [expr.const]/4 of the C standard:

• An expression is not a core constant expression if it evaluates an id-expression that refers to a variable or data member of reference type unless:

  • The reference has a preceding initialization and is usable in constant expressions
  • The lifetime of the reference began within the evaluation of the expression

In this case, the reference parameter data1 does not have a preceding initialization, so it cannot be used in the constant expression data1.size() data2.size().

Solution

To resolve the issue, simply replace data1.size() with the template parameter S1:

std::array<uint8_t s1 s2> result;</uint8_t>

The above is the detailed content of Why Doesn\'t `data1.size()` Work in a `constexpr` Function with Reference Parameters?. For more information, please follow other related articles on the PHP Chinese website!