how to write an overloaded with ignore feature while visit a variant

Question

Code:

#include <iostream>
#include <string>
#include <vector>
#include <variant>

struct ignore
{
    template <typename T>
    void operator()([[maybe_unused]]const T&)
    {
        std::cout << "other type" << std::endl;
    }
};

template <class... Ts>
struct overloaded_ignore : ignore, Ts...
{
    overloaded_ignore(Ts...) : ignore(){}
    using Ts::operator()...;
    using ignore::operator();
};

int main()
{
    std::variant<int,float,std::string> my_var = std::string("helloworld");
    // std::variant<int,float,std::string> my_var = 5.0F;
    // std::variant<int,float,std::string> my_var = 3;


    std::visit(overloaded_ignore{
        [](const int& t)
        {
            std::cout << "int var: " << t << std::endl;
        },
        [](const std::string& t)
        {
            std::cout << "string var: " << t << std::endl;
        }
        }, my_var);

    
    return 0;
}

I expect to output "string var: helloworld", but the "other type" is outputed.

How to fix this?

Note: the ignored operator() is required.

康桓瑋 · Accepted Answer · 2024-07-02 07:22:29Z

3

Just add the const-qualifier to ignore::operator() as the lambda's operator() is const-qualified by default:

struct ignore
{
    template <typename T>
    void operator()([[maybe_unused]]const T&) const // <--
    {
        std::cout << "other type" << std::endl;
    }
};

template <class... Ts>
struct overloaded_ignore : ignore, Ts...
{
    overloaded_ignore(Ts...) : ignore(){}
    using Ts::operator()...;
    using ignore::operator();
};

Demo

(Since overloaded_ignore{...} is a prvalue, this makes the non-const ignore::operator() a better match before)

edited Jul 2, 2024 at 7:22

answered Jul 2, 2024 at 7:03

康桓瑋

46.8k5 gold badges69 silver badges135 bronze badges

Sign up to request clarification or add additional context in comments.

6 Comments

JeJo Over a year ago

If you move the operator() to the overloaded_ignore, you can also save the constructor and one of the using ...: godbolt.org/z/nThvYaosr

anti-walker Over a year ago

it works. your really surprise me since you fix this issue by so easy way. thank you very much.

anti-walker Over a year ago

yes, actually at beginning i indeed leave the operator() in overloaded_ignore like you said but without const qualifier, so i can't get the correct output, so i move it to a new struct. i was thinking since the "Ts"(which is lambda in constructor argument list) is a fixed type, so it should be a better match than template<T> void ignore::operator()(auto&t); but it's not.

anti-walker Over a year ago

could you please explain why the ignore::operator() is a better match than lambda in my implementation, even the object of overloaded_ignore is a prvalue, but why compile choose ignore::operator() first? @JeJo

康桓瑋 Over a year ago

@anti-walker Because you are calling operator() of overloaded_ignore{...}, which is a prvalue value rather than a const object. If you create a const overloaded_ignore object, then the lambda will be a better match if you call its operator().

|

Collectives™ on Stack Overflow

how to write an overloaded with ignore feature while visit a variant

1 Answer 1

6 Comments

Your Answer

Hot Network Questions

Collectives™ on Stack Overflow

1 Answer 1

6 Comments

Your Answer

Sign up or log in

Post as a guest

Related