Different type in lamda with reference in capture list

It is fine to compile these code in c11

double(*p1)(double) = (double a) return sqrt(a); ;

double(*p2)(double) = [&](double a) return sqrt(a); ;


But there is the following error in compilation in c17

double(*p2)(double) = [&](double a) return sqrt(a); ;

cannot convert test_pointer_to_function::test_method()::<lambda(double)> to double (*)(double) in initialization double(*p2)(double) = [&](double a) return sqrt(a); ;

What is the correct type for this lamda function?

code blocking

std::function<double(double)> p2 = ...

auto p2 = ...

test_pointer_to_function

test_method

2 Answers
2

There's no "correct" function pointer type for lambda expression with non-empty lambda-capture. Such closure objects are not convertible to ordinary function pointers.

If you want to be able to convert closure object to ordinary function pointer type, make sure you have nothing in the part of lambda expression. This applies to C++11 and later, including C++17.

Each lambda has its own unique type. Normally, if you need to assign a lambda to a variable, you use auto for its type. If you need a known type (e.g. you are going to store lambdas in a container), you can use std::function with the same signature as your lambda. In your example, this will be

auto

std::function

std::function<double(double)> p2 = [&](double a) return sqrt(a); ;


(but note that you generally lose some efficiency this way).

Only non-capturing lambdas, i.e. lambdas with empty , are convertible to a function pointer, like in your example:

double(*p1)(double) = (double a) return sqrt(a); ;


Your C++11 compiler should not have accepted

double(*p2)(double) = [&](double a) return sqrt(a); ;


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