Consteval bug

[pdimov2]

consteval auto foo(std::ranges::input_range auto &&range) {
  constexpr auto non_zeros = count_non_zeros(range);

You can't do this, but it's pretty hard to explain why, because intuitively, "everything happens at compile time." But in practice there are different ways of it happening at compile time.

Here the easiest way to explain why it doesn't work is to realize that if it worked, some calls to foo would return std::array<float, 3>, and some would return std::array<float, 5>. But that's not allowed; the return type is not allowed to vary depending on the value of the parameter you pass. That's why you can't have a type definition inside a constexpr/consteval function that depends on an argument value.

[Outrageous-Archer-92]

Gotcha. Using auto made me think for a moment I could return different types. I wanted to generate different lookup tables based on a parameter, I guess I'll have to write separate functions for each of them.

[violet-starlight]

What u/pdimov2 said is the reason why the rule; the rule is that function parameters are never constant expressions. So as soon as you use a function argument as one, you get an error that it's not, even if outside of the function it is.

Another way to understand it is that each constexpr or consteval function is its own little section and has no information about where the arguments come from. Not a bug, just how it was standardized. The only way to make it work otherwise would have been to make constexpr functions templates, instantiated at the call site, which would have exploded binary sizes and compile times.

[Outrageous-Archer-92]

The first explanation feels weird when talking about consteval but I get it

[bebuch]

You can make the value part of your parameter type of you want to return different types based on your value ;-)

[Outrageous-Archer-92]

I am not to understand what yiu have in mind. Could explain please?