How can I initialize this std::variant data member which has two similar alternatives?

Question

The following code does not compile. Why is this restriction, and how can I bypass it? If I replace the variant declaration, it compiles.

struct PPP
{
    int xxx;
    PPP() : x(xxx) {} // error: No matching constructor for initialization of 'std::variant<Inner1, Inner2>'

    struct Inner1
    {
        Inner1(int &x )  : c(x) {}
        int &c;
    };
    struct Inner2
    {
        Inner2(int &x )  : c(x) {}
        int &c;
    };
    struct Inner3 {};

    std::variant<Inner1, Inner2> x;
    // std::variant<Inner1, Inner3> x; // This will cause the code to compile
};

Remy Lebeau · Accepted Answer · 2023-10-18 19:11:47Z

5

The xxx member is an int.

Inner1 and Inner2 are both constructable from (a reference to) an int, so x(xxx) is ambiguous for std::variant<Inner1, Inner2> as the compiler doesn't know which type you want to construct inside the std::variant.

When you use std::variant<Inner1, Inner3> instead, there is no ambiguity since Inner3 can't be constructed from an int.

UPDATE: As Raymond mentioned in a comment, in this case you can use one of the std::in_place_type_t or std::in_place_index_t constructors of std::variant to tell the compiler which inner type you want to create from the int.

edited Oct 18, 2023 at 19:11

answered Oct 18, 2023 at 18:59

Remy Lebeau

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2 Comments

Raymond Chen Over a year ago

Note that std::variant<Inner1, Inner2> is perfectly legal. But you have to give the constructor some help. x(std::in_place_type<Inner1>, xxx) or x(std::in_place_type<Inner2>, xxx), depending on which variant you want to initialize.

Raymond Chen Over a year ago

A simpler alternative is x(Inner1{xxx}). Note that this alternative creates a temporary Inner1 object that is moved into the variant. That's fine for this particular Inner1, but that might cause a problem for more complicated Inner1 implementations.

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