Question

I have a class db_interface. And defined a lambda type:

typedef void (*db_interface_lambda)();

When I create lambda in class in such way: [](){ /* do something */ }, it has good type (db_interface_lambda), but when I use [this](){ /* do something */ }, the compiler starts to shout at me.

cannot convert ‘db_interface::db_interface(std::ifstream&)::<lambda()>’ to ‘std::map<std::basic_string<char>, void (*)()>::mapped_type {aka void (*)()}’ in assignment

How to solve that problem? What is the correct type?

Answer 1

Because lambdas are only implicitly convertible to function pointers if and only if they do not capture anything.

§5.1.2 [expr.prim.lambda] p6

The closure type for a lambda-expression with no lambda-capture ([] is empty) has a public non-virtual non-explicit const conversion function to pointer to function having the same parameter and return types as the closure type’s function call operator.

Btw, what you typedef‘d there is a function pointer, not a lambda type. Lambda expressions have a unique, unnamed, nonunion class type. You can not name them.

§5.1.2 [expr.prim.lambda] p3

The type of the lambda-expression (which is also the type of the closure object) is a unique, unnamed nonunion class type