How to use lambda functions in C++11?

Question

I have an object that needs a function passed to it in order to work. Something like

int doSomething(string a, int b, string(*lookUpFunction)(string a, int b)) {
    return lookUpFunction(a,b).length():
}

This is simplified to just include needed parameters and correct return types. I believe I have defined this correctly but please correct me if I am wrong. lookUpFunction takes a string and an int and returns a string.

Where my problem comes in is using this function. I would think I would use like this

class x {
    DataBase* db;
public:
    int myFunc();
}


int x::myFunc() {
    return doSomething("a",1,[this](string a, int b)->string {
        return db->someFuncThatTakesStringAndInt(a,b);
    });
}

But my IDE throws errors. If I remove "this" it says db isn't accessible. With it in it says the lambda function is not castable to correct format.

You want the parameter to be a `std::function` rather than a function pointer. — HolyBlackCat, Jun 06 '23 at 19:15
One of the skills you need as a programmer is to read documentation. Have a look at [introduction to lambda functions learncpp.com](https://www.learncpp.com/cpp-tutorial/introduction-to-lambdas-anonymous-functions/) then you will need to learn about function templates. And you can write something like `template auto dosomething(function_type fn){fn();};` and call it with a lambda like this : `dosomething([]{ std::cout << "this is a lambda\n"; });`. Or in your case `dosomething(std::function` is also an option. — Pepijn Kramer, Jun 06 '23 at 19:16
See [std::function](https://en.cppreference.com/w/cpp/utility/functional/function) — Pepijn Kramer, Jun 06 '23 at 19:19
A lambda is convertible to a function pointer *only* if it doesn't capture anything. — BoP, Jun 06 '23 at 19:19
@BoP I was just going to say this too. So no do not try to convert lambdas to function pointers. Use a template or std::function. They are much more flexible. — Pepijn Kramer, Jun 06 '23 at 19:20
Thanks I will have to try both the template and std function approach and see which works better. Thanks again. — Matthew Cornelisse, Jun 06 '23 at 19:23
IMO question suffers form [XY problem](https://mywiki.wooledge.org/XyProblem). You should describe what code should do not how you try achieve that. — Marek R, Jun 06 '23 at 19:29

Remy Lebeau · Accepted Answer · 2023-06-06T19:25:15.557

Your doSomething() function is taking in a plain C-style function pointer for its lookUpFunction callback.

However, a capturing lambda is not convertible to a function pointer, only a non-capturing lambda is convertible. You are capturing this, which is why you can't pass in your lambda as shown.

Instead of a function pointer, your doSomething() function will need to take in a callback object as either a std::function or a template parameter, eg:

int doSomething(string a, int b, std::function<string(string, int)> lookUpFunction) {
    return lookUpFunction(a,b).length():
}

template<typename Callable>
int doSomething(string a, int b, Callable lookUpFunction) {
    return lookUpFunction(a,b).length():
}

Now, you can pass your capturing lambda to either one.

How to use lambda functions in C++11?

1 Answers1