c++ - c++ lambda表达式中的非法捕获

Question

我在为相当简单的 lambda 表达式编写有效捕获时遇到了一些困难。这是我要编译的代码:

#include <iostream>
#include <vector>

class State { public:
    int i;
    float f;
    State(int i,float f){this->i = i; this->f = f;}
};

typedef State (*FunctionType)(const State &state);


int main(int argc, char **argv)
{
    std::vector<FunctionType> funcs;

    funcs.push_back(
        [](const State &state)
        {
            return State(state.i+7,state.f-3.5);
        });

    State s(100,5.5);

    int m = 5;

    funcs.push_back(
        [](const State &state)
        // [=](const State &state)
        // [&](const State &state)
        {
            return State(m,m+0.5);
        });

    for (auto func : funcs)
    {
        std::cout << func(s).i << " " << func(s).f << "\n";
    }

    return 0;
}

当我编译它时

$ g++ -std=c++17 main.cpp -o main

我收到以下错误(以及更多错误)，表明我无法捕获 m:

main.cpp: In lambda function:
main.cpp:32:17: error: ‘m’ is not captured
    return State(m,m+0.5);
                 ^
main.cpp:28:4: note: the lambda has no capture-default
   [](const State &state)
    ^
main.cpp:25:6: note: ‘int m’ declared here
  int m = 5;
      ^

Answer 1

您确实需要您的评论版本之一来捕获 m

• [=](const State &state) { 返回状态(m, m + 0.5); }
• [&](const State &state) { 返回状态(m, m + 0.5); }

或显式捕获:

• [m](const State &state) { return State(m, m + 0.5); }
• [&m](const State &state) { 返回状态(m, m + 0.5); }

但是你有以下问题:

std::vector<FunctionType> funcs;

funcs.push_back([=](const State &state) { return State(m, m + 0.5); })

捕获lambda不能衰减到函数指针，需要改

typedef State (*FunctionType)(const State &state);

进入

using FunctionType = std::function<State(const State&)>;

可以处理lambda