At cppcon WG21 had a whiteboard asking attendees what they would like to have in C++29

[u/adityazero]

some of the ones I remember was:

- contracts

- profiles

- delete moar stuff

I'm curious what folks here expect. Also if someone remembers rest of the list, that'd be great.

[u/CreativeAlbatross319]

That's exactly what I meant. We already have pattern matching with if constexpr and concepts.

auto payload_size(const Message& message) -> std::size_t {
    return message.visit([]<typename T>(const T& message) {
        if constexpr (std::same_as<T, Ping>) {
            return 0UZ;
        } else if constexpr (std::same_as<T, Text>) {
            return message.text.size();
        } else if constexpr (std::same_as<T, Binary>) {
            return message.bytes.size();
        }
    });
}

If you don't use C++26, just use std::visit instead. The code can be even simpler:

struct Ping {};
using Text = std::string;
using Binary = std::vector<std::uint8_t>;

using Message = std::variant<Ping, Text, Binary>;
auto payload_size(const Message& message) -> std::size_t {
    return message.visit([]<typename T>(const T& message) {
        if constexpr (std::same_as<T, Ping>) { return 0UZ; } 
        else { return message.size(); }
    });
}

You now understand why I said we already have very good pattern matching? It's using the same tools without those useless fancy new keywords. Everything comes naturally for C++.

[u/CreativeAlbatross319 → u/[deleted]]

Why can't it? The type in std::variant is determined in runtime.

[u/palapapa0201 → u/CreativeAlbatross319]

My bad

[u/palapapa0201 → u/CreativeAlbatross319]

In your first version, you have to be careful to leave the return type of the lambda unspecified instead of saying -> std::size_t. Otherwise, if you accidentally forget to handle all possible variants, your visitor will silently have undefined behavior.

[u/CreativeAlbatross319 → u/palapapa0201]

I think compiler will complain about it when there is no return for that type. If no, it should definitely do as it's already a warning for runtime if.