Return variant from a lambda

Question

I have this simple lambda:

    std::variant<int, char> myLambda = []() { // no suitable user-defined conversion from "type" to "std::variant<int, char>" exists
        std::variant<int, char> res;

        if (true)
        {
            res = 1;
        }
        else
        {
            res = 'c'; 
        }

        return res;
    };

But it doesn't compile, producing error no suitable user-defined conversion from "type" to "std::variant<int, char>" exists. What am I doing wrong?

You need to call the lambda change the last `};` to `}();` Typo? — Richard Critten, Aug 30 '19 at 12:29
@RichardCritten No, I just want to initialize lambda, not to call it. — Nurbol Alpysbayev, Aug 30 '19 at 12:30
@RichardCritten However, you made me understand my mistake, thanks! — Nurbol Alpysbayev, Aug 30 '19 at 12:33
A problem with 2 solutions .... I suggested the wrong one. Glad the answers below helped you. — Richard Critten, Aug 30 '19 at 12:35
Possibly: `auto myLambda = [&]() -> std::variant { if (cond) { return 1; } else { return 'c'; }}};`. — Jarod42, Aug 30 '19 at 13:53
@Jarod42 Thanks! didn't know this way to define return type. Much cleaner! — Nurbol Alpysbayev, Aug 30 '19 at 14:02

score 7 · Answer 1 · edited Aug 30 '19 at 13:27

7

Either you mean

std::variant<int, char> v = []() {
    std::variant<int, char> res;

    if (true)
    {
        res = 1;
    }
    else
    {
        res = 'c'; 
    }

    return res;
}();
^^^

Or you mean

auto myLambda = []() {
    std::variant<int, char> res;

    if (true)
    {
        res = 1;
    }
    else
    {
        res = 'c'; 
    }

    return res;
};

Lambda expressions have unique types.

edited Aug 30 '19 at 13:27

Lightness Races in Orbit

378,754
76
643
1,055

answered Aug 30 '19 at 12:34

Vlad from Moscow

301,070
26
186
335

João Paulo · Accepted Answer · 2020-05-13T14:29:04.723

6

The lambda expression type is wrong. You're trying to bind to std::variant<int, char>. Lambda expressions type name is impl-defined. Use auto:

auto processProjectFile = []() {
    std::variant<int, char> res;
    if (true) {
        res = 1;
    } else {
        res = 'c'; 
    }
    return res;
};

Optionally, you can cast the lambda type to std::function replacing auto by std::function<std::variant<int, char>(void)>.

But if you intend to call the lambda, just replace }; at the end by }();.

edited May 13 '20 at 14:29

answered Aug 30 '19 at 12:30

João Paulo

6,300
4
51
80

1

Yep, this is what I've missed. Thanks. Stupid error I have had. BTW the `std::function` part is handy! Thanks! – Nurbol Alpysbayev Aug 30 '19 at 12:32
2

Not sure what you mean by `ClosureType`. The name of a type of a lambda is impl-defined – Lightness Races in Orbit Aug 30 '19 at 13:27

Return variant from a lambda

2 Answers2