Question

我很难为一个简单的lambda表达式编写有效的捕获。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;
}

当我用

编译时

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

我收到以下错误（以及更多），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

您确实需要一个注释版本来捕获m

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

或显式捕获：

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

但随后您遇到了以下问题：

std::vector<FunctionType> funcs;

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

捕获lambda无法衰减为函数指针，您需要更改

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

进入

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

可以处理lambda的

在C ++中Lambda表达式中的非法捕获

1 个答案: