CODEWITHSUNDEEP
c++pointerslambdaanonymous-functionauto
yanpas
yanpas

Reputation: 2225

How declare lambda with function pointers (without auto)?

I can declare anonymous functions (they are the same as lambda, but without "context" - [...]) without auto easily:

#include <iostream>

using namespace ::std;

void foo(void (*f)(char))
{
    f('r');
}

int main()
{
    void (*anon)(char) = [](char c) -> void { cout << c << endl; };
    foo(anon);
    return 0;
}

But how to declare lambda ? Is it the only possible way? (maybe with typedef). Here I use ::std::function, but I didn't mention context of f somewhere in foo arguments...:

#include <iostream>
#include <functional>

using namespace ::std;

//int foo(auto f, char x) // !only since c++14
int foo(function<int(char)> f, char x)
{
    return f(x+1);
}

int main()
{
    int a = 5, b = 10;

    //void (*lambda)(char) = [](char c) { cout << c << endl; };
    //auto sample_lambda = [&a,b](char c) -> int // !only since c++11
    function<int(char)> sample_lambda = [&a,b](char c) -> int  
        {
            for (;a<b;a++)
                cout << a << c << endl;
            return (int)c+a; 
        };

    foo(sample_lambda, 's');
    return 0;
}

Upvotes: 2

Views: 719

Answers (2)

Guillaume Racicot
Guillaume Racicot

Reputation: 41750

auto as parameter is an extension provided by GCC and others. What you can do instead is to use templates: 

template<typename T>
int callFoo(T funct, int x) {
    return funct(x+1);
}

Upvotes: 0

TartanLlama
TartanLlama

Reputation: 65600

You can't.

The type of a closure is a unique, unnamed type. Your first example works because closure types have a conversion function to a pointer-to-function if they don't capture anything, not because the closure is a void (*) (char).

You are best to just stick with auto as std::function can have additional overhead for the type erasure.

Upvotes: 5

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