Type conversion for lambda C++

Question

Is it possible to pass a lambda function as a function of some type?

For example, I have

typedef double(*Function)(int, double);

How can I convert a lambda function to the type?

See [Passing lambda as function pointer](https://stackoverflow.com/q/28746744/1708801) — Shafik Yaghmour, Sep 18 '17 at 19:23

score 7 · Answer 1 · answered Sep 18 '17 at 18:08

7

For a stateless lambda, this conversion is available implicitly:

Function f = [](int, double) -> double {};

If you just want to convert the lambda expression itself, you can use the unary + operator to achieve this:

auto* p = +[](int, double) -> double {};   // p is of type double(*)(int, double)

A lambda with a capture offers no such conversion, since it is not semantically equivalent to a (stateless) function (this applies even to the case of a default capture that doesn't end up capturing anything).

answered Sep 18 '17 at 18:08

Kerrek SB

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_"unary + operator"_ That's great! (As mentioned [here](https://stackoverflow.com/questions/46283731/a-function-definition-is-not-allowed-here-before-token#comment79533605_46283731) there are still white spots to explore on the maps). – user0042 Sep 18 '17 at 18:16
1

@user0042 - It is great. But sadly, a certain mainstream compiler, that shall remain unnamed, [still chokes on it](http://rextester.com/BQND42154). – StoryTeller - Unslander Monica Sep 18 '17 at 18:27
1

See https://stackoverflow.com/a/17822241/801894 for a deeper explanation of the unary + option. – Solomon Slow Sep 18 '17 at 19:22

score 2 · Accepted Answer · answered Sep 18 '17 at 19:28

You could use std::function instead of a plain function pointer. This would allow to hold lambdas even when they have captures:

#include <iostream>
#include <functional>

typedef std::function<double(int, double)> Function;

int main()
{
    Function f;
    int i = 4;
    f = [=](int l, double d){ return i*l*d; };
    std::cout << f(2, 3);
}

[Live Demo]

Type conversion for lambda C++

2 Answers2