Assign a lambda function to static member variable (c++)

Question

I have a template class with a static member variable that I want to be a lambda function. Why doesn't it work?

#include <iostream>
using namespace std;

template <typename T>
class Test {
public:
    static constexpr auto lambda = [](T val) -> T {
        return val;
    };
};

int main() {
    cout << Test<int>::lambda(123) << endl;
}

I tried both version with constexpr and const.

In the first case I get:

../src/test.cpp:8:24: error: ‘constexpr const Test<int>::<lambda(int)> Test<int>::lambda’, declared using local type ‘const Test<int>::<lambda(int)>’, is used but never defined [-fpermissive]
  static constexpr auto lambda = [](T val) -> T {
                        ^~~~~~

In the second case:

../src/test.cpp:7:20: error: ‘constexpr’ needed for in-class initialization of static data member ‘const Test<int>::<lambda(int)> Test<int>::lambda’ of non-integral type [-fpermissive]
  static const auto lambda = [](T val) -> T {
                    ^~~~~~
../src/test.cpp:7:20: error: ‘const Test<int>::<lambda(int)> Test<int>::lambda’, declared using local type ‘const Test<int>::<lambda(int)>’, is used but never defined [-fpermissive]

Why don't you simply use a static function? ...after all there will be different of such for each separate instantiation of Test.. — WhiZTiM
– WhiZTiM, Commented Mar 25, 2017 at 10:03
Because I need to refer to the function with a very short name, like n(arg) and variable assignment is a very intuitive concept. Instead, by using a static function the user should define a reference n to Test<int>::myFunction. I didn't even know function reference existed until now. — barsdeveloper
– barsdeveloper, Commented Mar 25, 2017 at 10:07

DaveFar · Accepted Answer · 2017-03-25 11:02:57Z

2

A core constant expression may contain a lambda only starting from C++17 (see cppreference point 8). This was proposal N24487 and got into C++17 as P0170R0.

If you have to use a static lambda, you can make use of the construct at first use idiom:

#include <iostream>

template <typename T>
class Test {
public:
    std::function<T(T)> createLambda() {
        static const std::function<T(T)> returnLambda = [](T val) -> T {
            return val;
        };
        return returnLambda;
    }
};

int main() {
    Test<int> lambdaFactory;
    std::function<int(int)> n = lambdaFactory.createLambda();
    std::cout << n(123) << std::endl;
}

edited Mar 25, 2017 at 11:02

answered Mar 25, 2017 at 10:31

DaveFar

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1 Comment

Sebastian Redl Over a year ago

If you've got C++14 deduced return types, you can probably get rid of the std::function wrapper.

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