问题
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 ?
回答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]
).
回答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.
来源:https://stackoverflow.com/questions/13461538/lambda-of-a-lambda-the-function-is-not-captured