C++ lambda 的 Lambda ：函数未被捕获
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Lambda of a lambda : the function is not captured
提问by Vincent
The following program do not compile :
以下程序无法编译：
#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 :
因为 ：
error : 'f' is not captured
What does it mean and how to solve the problem ?
这是什么意思以及如何解决问题？
回答by cdhowie
You use the fparameter in the lambda inside asort(), but you don't capture it. Try adding fto the capture list (change []to read [&f]).
您f在 lambda 内部使用参数asort()，但您没有捕获它。尝试添加f到捕获列表（更改[]为 read [&f]）。
回答by wendazhou
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).
您在 lambda 中有效地引用了 f，它是外部作用域中的一个变量。您应该在您的捕获列表中捕获它（最简单的可能是通过引用 [&f] 或 [&] 通过引用捕获所有内容，因为您正在立即使用它）。
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.
另一方面， std::function 在执行类型擦除时有一些开销，在您的情况下，最好引入模板类型。
