Lambda of a lambda : the function is not captured

Question

The following program do not compile :

#include <iostream>
#include <vector>
#include <functional>
#include <algorithm>
#include <cstdlib>
#include <cmath>

void asort(std::vector<double>& v, std::function<bool(double, double)> f)
{
    std::sort(v.begin(), v.end(), [](double a, double b){return f(std::abs(a), std::abs(b));});
}

int main()
{
    std::vector<double> v({1.2, -1.3, 4.5, 2.3, -10.2, -3.4});
    for (unsigned int i = 0; i < v.size(); ++i) {
        std::cout<<v[i]<<" ";
    }
    std::cout<<std::endl;
    asort(v, [](double a, double b){return a < b;});
    for (unsigned int i = 0; i < v.size(); ++i) {
        std::cout<<v[i]<<" ";
    }
    std::cout<<std::endl;
    return 0;
}

because :

error : 'f' is not captured

What does it mean and how to solve the problem ?

Solution 1

You use the f parameter in the lambda inside asort(), but you don't capture it. Try adding f to the capture list (change [] to read [&f]).

Solution 2

You are effectively referencing f, which is a variable in the outer scope, in your lambda. You should capture it in your capture list (simplest is probably by reference [&f], or [&] to capture everything by reference, as you are using it immediately).

On another note, std::function has some overhead as it performs type erasure, in your case here it might be better to introduce a template type.

Comments

[Cameron]

Or even capture it by reference with [&].

[cdhowie]

Yes, you are right, I've updated my answer. For some reason I read too fast and thought that f was a function pointer, not a function object.